Selecting the smallest two elements on the whiteboard is a good choice in the first move.

Let $$$b$$$ denote the sorted array $$$a$$$. Assume that $$$b$$$ contains only distinct elements for convenience. We prove by induction on $$$n$$$ that the maximum final score is $$$b_1 + b_3 + \ldots + b_{2n-1}$$$.

For the base case $$$n = 1$$$, the final and only possible score that can be achieved is $$$b_1$$$.

Now let $$$n \gt 1$$$.

Claim: It is optimal to choose $$$b_{1}$$$ with $$$b_{2}$$$ for some move.

Note that we can extend the arguments for the case where $$$a$$$ has duplicate elements.

#include <bits/stdc++.h>     
using namespace std;
#define ll long long
void solve(){ 
    ll n; cin>>n;
    vector<ll> a(2*n);
    ll ans=0;
    for(auto &it:a){
        cin>>it;
    }
    sort(a.begin(),a.end());
    for(ll i=0;i<2*n;i+=2){
        ans+=a[i];
    }
    cout<<ans<<"\n";
    return;  
}                                       
int main()                                                                               
{       
    ios_base::sync_with_stdio(false);                         
    cin.tie(NULL);                               
    #ifndef ONLINE_JUDGE                   
    freopen("input.txt", "r", stdin);                                           
    freopen("output.txt", "w", stdout);      
    freopen("error.txt", "w", stderr);                        
    #endif     
    ll test_cases=1;                 
    cin>>test_cases;
    while(test_cases--){
        solve();
    }
    cout<<fixed<<setprecision(10);
    cerr<<"Time:"<<1000*((double)clock())/(double)CLOCKS_PER_SEC<<"ms\n"; 
}  

For integers $$$x$$$ ($$$\lfloor \frac{n}{2} \rfloor \lt x \leq n$$$), there does not exist integer $$$y$$$ ($$$y \gt x$$$) such that $$$y$$$ is divisible by $$$x$$$.

Consider the permutation $$$p$$$ such that $$$p=[1, n, 2, n - 1, \ldots \lceil \frac{n+1}{2} \rceil]$$$. It is valid. Why?

#include <bits/stdc++.h>     
using namespace std;
#define ll long long
void solve(){ 
    ll n; cin>>n;
    ll l=1,r=n;
    for(ll i=1;i<=n;i++){
        if(i&1){
            cout<<l<<" ";
            l++;
        }
        else{
            cout<<r<<" ";
            r--;
        }
    }
    cout<<"\n";
    return;  
}                                       
int main()                                                                               
{       
    ios_base::sync_with_stdio(false);                         
    cin.tie(NULL);                               
    #ifndef ONLINE_JUDGE                   
    freopen("input.txt", "r", stdin);                                           
    freopen("output.txt", "w", stdout);      
    freopen("error.txt", "w", stderr);                        
    #endif     
    ll test_cases=1;                 
    cin>>test_cases;
    while(test_cases--){
        solve();
    }
    cout<<fixed<<setprecision(10);
    cerr<<"Time:"<<1000*((double)clock())/(double)CLOCKS_PER_SEC<<"ms\n"; 
}  

Consider an array $$$c$$$ of length $$$n$$$ such that $$$c_i :=$$$ number of indices smaller than $$$i$$$ which were chosen before index $$$i$$$.

So set $$$S$$$ will be a collection of $$$a_i + i - c_i$$$ over all $$$1 \leq i \leq n$$$.

Now one might wonder what type of arrays $$$c$$$ is it possible to get.

First, it is easy to see that we should have $$$0 \leq c_i \lt i$$$ for all $$$i$$$. Call an array $$$c$$$ of length $$$n$$$ good, if $$$0 \leq c_i \lt i$$$ for all $$$1 \leq i \leq n$$$. The claim is that all good arrays of length $$$n$$$ can be obtained.

So we have the following subproblem. We have a set $$$S$$$. We will iterate $$$i$$$ from $$$1$$$ to $$$n$$$, select an integer $$$c_i$$$ ($$$0 \leq c_i \leq i - 1$$$) and insert $$$a_i + i - c_i$$$ into set $$$S$$$ and move to $$$i + 1$$$. Now using exchange arguments, we can prove that it is never bad if we select the smallest integer $$$v$$$ ($$$0 \leq v \leq i - 1$$$) such that $$$a_i + i - v$$$ is not present in the set $$$S$$$, and assign it to $$$c_i$$$. Note that as we have $$$i$$$ options for $$$v$$$, and we would have inserted exactly $$$i-1$$$ elements before index $$$i$$$, there always exists an integer $$$v$$$ ($$$0 \leq v \leq i - 1$$$) such that $$$a_i + i - v$$$ is not present in the set $$$S$$$. You can refer to the attached submission to see how to find $$$v$$$ efficiently for each $$$i$$$.

#include <bits/stdc++.h>     
using namespace std;
#define ll long long
void solve(){ 
    ll n; cin>>n;
    set<ll> used,not_used;
    vector<ll> ans;
    for(ll i=1;i<=n;i++){
        ll x; cin>>x; x+=i;
        if(!used.count(x)){
            not_used.insert(x);
        }
        ll cur=*(--not_used.upper_bound(x)); //find the largest element(<= x) which is not in set "used"
        not_used.erase(cur);
        ans.push_back(cur);
        used.insert(cur);
        if(!used.count(cur-1)){
            not_used.insert(cur-1);
        }
    }
    sort(ans.begin(), ans.end());
    reverse(ans.begin(), ans.end());
    for(auto i:ans){
        cout<<i<<" ";
    }  
    cout<<"\n";
    return;  
}                                       
int main()                                                                               
{       
    ios_base::sync_with_stdio(false);                         
    cin.tie(NULL);                               
    #ifndef ONLINE_JUDGE                   
    freopen("input.txt", "r", stdin);                                           
    freopen("output.txt", "w", stdout);      
    freopen("error.txt", "w", stderr);                        
    #endif     
    ll test_cases=1;                 
    cin>>test_cases;
    while(test_cases--){
        solve();
    }
    cout<<fixed<<setprecision(10);
    cerr<<"Time:"<<1000*((double)clock())/(double)CLOCKS_PER_SEC<<"ms\n"; 
}  

To find $$$f(s)$$$, we can partition $$$s$$$ into multiple independent substrings of length atmost $$$3$$$ and find best answer for them separately.

There always exists a string $$$g$$$ such that:

1. $$$g$$$ is $$$s-good$$$
2. there are $$$f(s)$$$ number of $$$\mathtt{1} $$$s in $$$g$$$
3. $$$g$$$ is of the form $$$b_1 + b_2 + \ldots b_q$$$, where $$$b_i$$$ is either equal to $$$\mathtt{0}$$$ or $$$\mathtt{010}$$$.

First of all, append $$$n$$$ $$$\mathtt{0} $$$ s to the back of $$$s$$$ for our convenience. Note that this does not change the answer.

Now let us call a binary string $$$p$$$ of size $$$d$$$ nice if:

1. there exists a positive integer $$$k$$$ such that $$$d = 3k$$$

2. $$$p$$$ is of form $$$ f(\mathtt{0} , k) + f(\mathtt{1} , k) + f(\mathtt{0} , k) $$$, where $$$f(c, z)$$$ gives a string containing exactly $$$z$$$ characters equal to $$$c$$$.

Suppose binary string $$$t$$$ is one of the $$$s-good$$$ strings such that there are exactly $$$f(s)$$$ $$$\mathtt{1} $$$ s in $$$t$$$.

We claim that for any valid $$$t$$$, there always exists a binary string $$$t'$$$ such that:

1. $$$t'$$$ is permutation of $$$t$$$
2. $$$t'$$$ is $$$s-good$$$
3. $$$t'$$$ is of the form $$$f(\mathtt{0}, d_1) + z_1 + f(\mathtt{0}, d_2) + z_2 + f(\mathtt{0}, d_3) + \ldots + z_g + f(\mathtt{0}, d_{g+1}) $$$, where $$$z_1, z_2, \ldots z_g$$$ are nice binary strings and $$$d_1, d_2, \ldots d_{g+1}$$$ are non-negative integers.

Now, carefully observe the structure of $$$t'$$$. We can replace all the substrings of the form $$$ f(\mathtt{0} , k) + f(\mathtt{1} , k) + f(\mathtt{0} , k) $$$ in $$$t'$$$ with $$$ \mathtt{0} \mathtt{1} \mathtt{0} \mathtt{0} \mathtt{1} \mathtt{0} \ldots \mathtt{0} \mathtt{1} \mathtt{0} \mathtt{0} \mathtt{1} \mathtt{0}$$$.

So the updated $$$t'$$$(say $$$t"$$$) will be of form $$$b_1 + b_2 + \ldots b_q$$$, where $$$b_i$$$ is either equal to $$$\mathtt{0}$$$ or $$$\mathtt{010}$$$.

So whenever we need to find $$$f(e)$$$ for some binary string $$$e$$$, we can always try to find a string of form $$$t"$$$ using as few $$$\mathtt{1} $$$s as possible. Notice that we can construct $$$t"$$$ greedily. You can look at the attached code for the implementation details. Also, we don't need to actually append the $$$\mathtt{0} $$$s at the back of $$$s$$$. It was just for proof purposes.

#include <bits/stdc++.h>   
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;   
using namespace std;
#define ll long long
#define pb push_back                  
#define mp make_pair          
#define nline "\n"                            
#define f first                                            
#define s second                                             
#define pll pair<ll,ll> 
#define all(x) x.begin(),x.end()     
const ll MOD=1e9+7;
const ll MAX=500500;
ll f(string s){
    ll len=s.size(),ans=0,pos=0;
    while(pos<len){
        if(s[pos]=='1'){
            ans++;
            pos+=2;
        }
        pos++;
    }
    return ans;
}
void solve(){ 
    ll n,ans=0; cin>>n;
    string s; cin>>s; 
    for(ll i=0;i<n;i++){
        string t; 
        for(ll j=i;j<n;j++){
            t.push_back(s[j]);
            ans+=f(t);
        }
    }
    cout<<ans<<nline;
    return;  
}                                       
int main()                                                                               
{     
    ios_base::sync_with_stdio(false);                         
    cin.tie(NULL);                               
    #ifndef ONLINE_JUDGE                 
    freopen("input.txt", "r", stdin);                                           
    freopen("output.txt", "w", stdout);      
    freopen("error.txt", "w", stderr);                        
    #endif     
    ll test_cases=1;                 
    cin>>test_cases;
    while(test_cases--){
        solve();
    }
    cout<<fixed<<setprecision(10);
    cerr<<"Time:"<<1000*((double)clock())/(double)CLOCKS_PER_SEC<<"ms\n"; 
}  

We can use the idea of D1 and dynamic programming to solve in $$$O(n)$$$.

Suppose $$$dp[i][j]$$$ denotes $$$f(s[i,j])$$$ for all $$$1 \le i \le j \le n$$$.

Performing the transition is quite easy.

If $$$s_i = \mathtt{1}$$$, $$$dp[i][j]=1+dp[i+3][j]$$$, otherwise $$$dp[i][j]=dp[i+1][j]$$$. Note that $$$dp[i][j] = 0$$$ for if $$$i \gt j$$$.

So if we fix $$$j$$$, we can find $$$dp[i][j]$$$ for all $$$1 \le i \le j$$$ in $$$O(n)$$$, and the original problem in $$$O(n^2)$$$.

Now, we need to optimise it.

Suppose $$$track[i] = \sum_{j=i}^{n} dp[i][j]$$$ for all $$$1 \le i \le n$$$, with base condition that $$$track[i] = 0$$$ if $$$i \gt n$$$.

There are two cases:

• $$$s_i = \mathtt{0}$$$ :

1. $$$track[i] = \sum_{j=i}^{n} dp[i][j]$$$
2. $$$track[i] = dp[i][i] + \sum_{j=i+1}^{n} dp[i][j]$$$
3. $$$track[i] = dp[i][i] + \sum_{j=i+1}^{n} dp[i+1][j]$$$
4. $$$track[i] = track[i+1]$$$ as $$$dp[i][i]=0$$$

• $$$s_i = \mathtt{1}$$$ :

1. $$$track[i] = \sum_{j=i}^{n} dp[i][j]$$$
2. $$$track[i] = \sum_{j=i}^{n} 1 + dp[i+3][j]$$$
3. $$$track[i] = n - i + 1 + \sum_{j=i+3}^{n} dp[i+3][j]$$$ as $$$dp[i+3][i]=dp[i+3][i+1]=dp[i+3][i+2]=0$$$
4. $$$track[i] = n - i + 1 + \sum_{j=i+3}^{n} dp[i+3][j]$$$
5. $$$track[i] = n - i + 1 + track[i+3]$$$

So, the answer to the original problem is $$$\sum_{i=1}^{n} track[i]$$$, which we can do in $$$O(n)$$$.

#include <bits/stdc++.h>   
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;   
using namespace std;
#define ll long long
#define pb push_back                  
#define mp make_pair          
#define nline "\n"                            
#define f first                                            
#define s second                                             
#define pll pair<ll,ll> 
#define all(x) x.begin(),x.end()     
const ll MOD=1e9+7;
const ll MAX=500500;
void solve(){ 
    ll n,ans=0; cin>>n;
    string s; cin>>s; s=" "+s;
    vector<ll> dp(n+5,0);
    for(ll i=n;i>=1;i--){
        if(s[i]=='1'){
            dp[i]=dp[i+3]+n-i+1;
        }
        else{
            dp[i]=dp[i+1];
        }
        ans+=dp[i];
    }
    cout<<ans<<nline;
    return;  
}                                       
int main()                                                                               
{     
    ios_base::sync_with_stdio(false);                         
    cin.tie(NULL);                               
    #ifndef ONLINE_JUDGE                 
    freopen("input.txt", "r", stdin);                                           
    freopen("output.txt", "w", stdout);      
    freopen("error.txt", "w", stderr);                        
    #endif     
    ll test_cases=1;                 
    cin>>test_cases;
    while(test_cases--){
        solve();
    }
    cout<<fixed<<setprecision(10);
    cerr<<"Time:"<<1000*((double)clock())/(double)CLOCKS_PER_SEC<<"ms\n"; 
}  

Suppose you are given some array $$$b$$$ of length $$$m$$$ and a positive integer $$$k$$$. How to check whether we can get the array $$$b$$$ if we start with an array $$$a$$$ of length $$$n$$$ such that $$$a_i = i$$$ for all $$$i$$$ ($$$1 \leq i \leq n$$$)?

First of all, array $$$b$$$ should be a subsequence of $$$a$$$. Now consider an increasing array $$$c$$$ (possibly empty) such that it contains all the elements of $$$a$$$ which are not present in $$$b$$$. Now look at some trivial necessary conditions.

• The length of array $$$c$$$ should divisible by $$$2k$$$, as exactly $$$2k$$$ elements were deleted in one operation.

• There should be atleast one element $$$v$$$ in $$$b$$$ such that there are atleast $$$k$$$ elements smaller than $$$v$$$ in the array $$$c$$$, and alteast $$$k$$$ elements greater than $$$v$$$ in the array $$$c$$$. Why? Think about the last operation. We can consider the case of empty $$$c$$$ separately.

In fact, it turns out that these necessary conditions are sufficient (Why?).

Now, we need to find the number of possible $$$b$$$. We can instead find the number of binary strings $$$s$$$ of length $$$n$$$ such that $$$s_i = 1$$$ if $$$i$$$ is present in $$$b$$$, and $$$s_i=0$$$ otherwise.

For given $$$n$$$ and $$$k$$$, let us call $$$s$$$ good if there exists some $$$b$$$ which can be achieved from $$$a$$$.

Instead of counting strings $$$s$$$ which are good, let us count the number of strings which are not good.

For convenience, we will only consider strings $$$s$$$ having the number of $$$\mathtt{0}$$$'s divisible by $$$2k$$$.

Now, based on the conditions in hint $$$2$$$, we can conclude that $$$s$$$ is bad if and only if there does not exist any $$$1$$$ between the $$$k$$$-th $$$0$$$ from the left and the $$$k$$$-th $$$0$$$ from the right in $$$s$$$.

Let us compress all the $$$\mathtt{0}$$$'s between the $$$k$$$-th $$$0$$$ from the left and the $$$k$$$-th $$$0$$$ from the right into a single $$$0$$$ and call the new string $$$t$$$. Note that $$$t$$$ will have exactly $$$2k - 1$$$ $$$\mathtt{0}$$$'s. We can also observe that for each $$$t$$$, a unique $$$s$$$ exists. This is only because we have already fixed the parameters $$$n$$$ and $$$k$$$. Thus the number of bad $$$s$$$ having exactly $$$x$$$ $$$\mathtt{1}$$$'s is $$${{x + 2k - 1} \choose {2k - 1}}$$$ as there are exactly $$${{x + 2k - 1} \choose {2k - 1}}$$$ binary strings $$$t$$$ having $$$2k - 1$$$ $$$\mathtt{0}$$$'s and $$$x$$$ $$$\mathtt{1}$$$'s.

Finally, there are exactly $$$\binom{n}{x} - \binom{x + 2k - 1}{2k - 1}$$$ good binary strings $$$s$$$ having $$$x$$$ $$$\mathtt{1}$$$'s and $$$n-x$$$ $$$\mathtt{0}$$$'s. Now, do we need to find this value for each $$$x$$$ from $$$1$$$ to $$$n$$$? No, as the number($$$n-x$$$) of $$$\mathtt{0}$$$'s in $$$s$$$ should be a multiple of $$$2k$$$. There are only $$$O(\frac{n}{2k})$$$ useful candidates for $$$x$$$.

Thus, our overall complexity is $$$O(n \log(n))$$$ (as $$$\sum_{i=1}^{n} O(\frac{n}{i}) = O(n \log(n))$$$).

#pragma GCC optimize("O3,unroll-loops")
#include <bits/stdc++.h>   
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;   
using namespace std;
#define ll long long
const ll INF_ADD=1e18;
#define pb push_back                  
#define mp make_pair          
#define nline "\n"                            
#define f first                                            
#define s second                                             
#define pll pair<ll,ll> 
#define all(x) x.begin(),x.end()     
const ll MOD=998244353;
const ll MAX=5000500;
vector<ll> fact(MAX+2,1),inv_fact(MAX+2,1);
ll binpow(ll a,ll b,ll MOD){
    ll ans=1;
    a%=MOD;  
    while(b){
        if(b&1)
            ans=(ans*a)%MOD;
        b/=2;
        a=(a*a)%MOD;
    }
    return ans;
}
ll inverse(ll a,ll MOD){
    return binpow(a,MOD-2,MOD);
} 
void precompute(ll MOD){
    for(ll i=2;i<MAX;i++){
        fact[i]=(fact[i-1]*i)%MOD;
    }
    inv_fact[MAX-1]=inverse(fact[MAX-1],MOD);
    for(ll i=MAX-2;i>=0;i--){
        inv_fact[i]=(inv_fact[i+1]*(i+1))%MOD;
    }
}
ll nCr(ll a,ll b,ll MOD){
    if(a==b){
        return 1;
    }
    if((a<0)||(a<b)||(b<0))
        return 0;   
    ll denom=(inv_fact[b]*inv_fact[a-b])%MOD; 
    return (denom*fact[a])%MOD;    
}
ll n,k;    
ll ways(ll gaps,ll options){
    gaps--;
    ll now=nCr(gaps+options-1,options-1,MOD);
    return now;
}
void solve(){
    cin>>n; 
    for(ll k=1;k<=(n-1)/2;k++){
        ll ans=1;
        for(ll deleted=2*k;deleted<=n-1;deleted+=2*k){
            ll options=2*k,left_elements=n-deleted;
            ans=(ans+ways(left_elements+1,deleted+1)-ways(left_elements+1,options)+MOD)%MOD;  
        }
        cout<<ans<<" ";
    }
    cout<<nline;
    return;   
}                                       
int main()                                                                               
{     
    ios_base::sync_with_stdio(false);                         
    cin.tie(NULL); 
    #ifndef ONLINE_JUDGE                 
    freopen("input.txt", "r", stdin);                                           
    freopen("output.txt", "w", stdout);      
    freopen("error.txt", "w", stderr);                        
    #endif  
    ll test_cases=1;                 
    cin>>test_cases;
    precompute(MOD);
    while(test_cases--){
        solve();
    }
    cout<<fixed<<setprecision(10);
    cerr<<"Time:"<<1000*((double)clock())/(double)CLOCKS_PER_SEC<<"ms\n"; 
}  

For an array $$$b$$$ consiting of $$$m$$$ non-negative integers, $$$f(b) = \max\limits_{p=1}^{m} ( \max\limits_{i = 1}^{m} (b_i | b_p) - \min\limits_{i = 1}^{m} (b_i | b_p))$$$. In other, we can get the maximum possible value by choosing $$$x=b_p$$$ for some $$$p$$$ ($$$1 \leq p \leq m$$$).

$$$f(b)$$$ is the maximum value of $$$b_i \land $$$ ~ $$$b_j$$$ over all pairs of ($$$i,j$$$) ($$$1 \leq i,j \leq m$$$), where $$$\land$$$ is the bitwise AND operator, and ~ is the bitwise One's complement operator.

Let us see how to find $$$f(b)$$$ in $$$O(n \log(n))$$$ first. We will focus on updates later. Let us have two sets $$$S_1$$$ and $$$S_2$$$ such that

• $$$S_1$$$ contains all submasks of $$$b_i$$$ for all $$$1 \leq i \leq m$$$
• $$$S_2$$$ contains all submasks of ~$$$b_i$$$for all $$$1 \leq i \leq m$$$

We can observe that $$$f(b)$$$ is the largest element present in both sets $$$S_1$$$ and $$$S_2$$$.

Now, we can insert all submasks naively. But it would be pretty inefficient. Note that we need to insert any submask atmost once in either of the sets.

Can you think of some approach in which you efficiently insert all the non-visited submasks of some mask?

insert_submask(cur, S){
    return if mask cur is present in S
    add mask cur to the set S
    vec = list of all the set bits of the mask cur
    for i in vec:
        insert_submask(cur - 2^i)
}

Note that the above pseudo code inserts all the all submasks efficiently. As all the masks will be visited atmost once, the amortised complexity will be $$$O(n \log(n)^2)$$$. Note that instead of using a set, we can use a boolean array of size $$$n$$$ to reduce the complexity to $$$O(n \log(n))$$$.

Thus, we can use the above idea to find $$$f(b)$$$ in $$$O(n \log(n))$$$. For each $$$i$$$ from $$$1$$$ to $$$m$$$, we can insert all submasks of $$$b_i$$$ into set $$$S_1$$$ and insert all the submasks of ~$$$b_i$$$ into set $$$S_2$$$.

The above idea hints at how to deal with updates. If we need to append an element $$$z$$$ to $$$b$$$, we can just insert all submasks of $$$z$$$ into set $$$S_1$$$ and insert all the submasks of ~$$$z$$$ into set $$$S_2$$$.

Hence, the overall complexity is $$$O(n \log(n))$$$.

#pragma GCC optimize("O3,unroll-loops")
#include <bits/stdc++.h>   
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;   
using namespace std;
#define ll long long
#define pb push_back                  
#define mp make_pair          
#define nline "\n"                            
#define f first                                            
#define s second                                             
#define pll pair<ll,ll> 
#define all(x) x.begin(),x.end() 
const ll MAX=100100; 
void solve(){     
    ll n,q; cin>>n>>q;
    ll till=1,len=1;
    while(till<n){
        till*=2;
        len++;
    }
    ll ans=0;
    vector<vector<ll>> track(2,vector<ll>(till+5,0));
    auto add=[&](ll x,ll p){
        queue<ll> trav;
        if(track[p][x]){
            return;
        }
        trav.push(x); track[p][x]=1;
        while(!trav.empty()){
            auto it=trav.front();
            trav.pop();
            if(track[0][it]&track[1][it]){
                ans=max(ans,it);
            }
            for(ll j=0;j<len;j++){
                if(it&(1<<j)){
                    ll cur=(it^(1<<j));   
                    if(!track[p][cur]){
                        track[p][cur]=1;
                        trav.push(cur);
                    }
                }
            }
        }
    };
    ll supermask=till-1;
    vector<ll> a(q+5);
    for(ll i=1;i<=q;i++){
        ll h; cin>>h; 
        a[i]=(h+ans)%n;
        add(a[i],0);
        add(supermask^a[i],1);
        cout<<ans<<" \n"[i==q];
    }
    return;  
}                                         
int main()                                                                               
{     
    ios_base::sync_with_stdio(false);                         
    cin.tie(NULL);                               
    #ifndef ONLINE_JUDGE                 
    freopen("input.txt", "r", stdin);                                           
    freopen("output.txt", "w", stdout);      
    freopen("error.txt", "w", stderr);                        
    #endif     
    ll test_cases=1;                 
    cin>>test_cases;
    while(test_cases--){
        solve();
    }
    cout<<fixed<<setprecision(10);
    cerr<<"Time:"<<1000*((double)clock())/(double)CLOCKS_PER_SEC<<"ms\n"; 
}  

Consider some subsequence $$$S$$$ of $$$[1,2, \ldots n]$$$ such that there exists atleast one pre-order $$$a$$$ such that $$$F(a)=S$$$. Look for some non-trivial properties about $$$S$$$. Node $$$S_i$$$ will be visited before the node $$$S_{i+1}$$$.

Assume $$$|S|=k$$$. First of all, we should $$$S_1=1$$$ and $$$S_k = n$$$. There cannot exists there distinct indices $$$a, b$$$ and $$$c$$$ ($$$1 \leq a \lt b \lt c \leq |k|$$$) such that $$$S_c$$$ lies on the path from $$$S_a$$$ to $$$S_b$$$. Assume $$$LCA_i$$$ is the lowest common ancestor of $$$S_i$$$ and $$$S_{i+1}$$$. For all $$$i$$$($$$1 \leq i \lt k$$$), the largest value over all the nodes on the unique path from $$$S_i$$$ to $$$S_{i+1}$$$ should be $$$S_{i+1}$$$.

Suppose $$$nax_p$$$ is the maximum value in the subtree of $$$p$$$. There is one more important restriction if $$$S_i$$$ does not lie on the path from $$$S_{i+1}$$$ to $$$1$$$. Suppose $$$m$$$ is the child of $$$LCA_i$$$, which lies on the path from $$$S_i$$$ to $$$LCA_i$$$. We should have $$$S_{i+1} \gt nax_m$$$. In fact, if you observe carefully you will realise that we should have $$$S_i = nax_m$$$.

Let us use dynamic programming. Suppose $$$dp[i]$$$ gives the number of valid subsequences(say $$$S$$$) such that the last element of $$$S$$$ is $$$i$$$. Note that the answer to our original problem will be $$$dp[n]$$$.

Suppose $$$nax(u,v)$$$ denotes the maximum value on the path from $$$u$$$ to $$$v$$$(including the endpoints).

Let us have a $$$O(n^2)$$$ solution first.

We have $$$dp[1]=1$$$.

Suppose we are at some node $$$i$$$($$$i \gt 1$$$), and we need to find $$$dp[i]$$$. Let us consider some node $$$j$$$($$$1 \le j \lt i$$$) and see if we can append $$$i$$$ to all the subsequences which end with node $$$j$$$. If we can append, we just need to add $$$dp[j]$$$ to $$$dp[i]$$$.

But how do we check if we can append $$$i$$$ to all the subsequences that end with node $$$j$$$?

Check hints 2 and 3.

So, we have a $$$n^2$$$ solution now.

We need to optimise it now.

We will move in the increasing value of the nodes(from $$$2$$$ to $$$n$$$) and calculate the $$$dp$$$ values.

Suppose $$$nax(1, par_i) = v$$$, where $$$par_i$$$ denotes the parent of $$$i$$$.

Assume we go from node $$$j$$$($$$j \lt i$$$) to node $$$i$$$.

There are two cases:

1. $$$j$$$ lies on the path from $$$1$$$ to $$$i$$$: This case is easy, as we just need to ensure that $$$nax(j,par_i) = j$$$. We can add $$$dp[j]$$$ to $$$dp[i]$$$ if we have $$$nax(j,par_i) = j$$$. Note that there exists only one node(node $$$v$$$) for which might add $$$dp[v]$$$ to $$$dp[i]$$$

2. $$$j$$$ does not lie on the path from $$$1$$$ to $$$i$$$ : We will only consider the case in which $$$dp[j]$$$ will be added to $$$dp[i]$$$. Suppose $$$lca$$$ is the lowest common ancestor of $$$j$$$ and $$$i$$$, and $$$m$$$ is the child of $$$lca$$$, which lies on the path from $$$j$$$ to $$$lca$$$. Notice that the largest value in the subtree of $$$m$$$ should be $$$j$$$. This observation is quite helpful. We can traverse over all the ancestors of $$$i$$$. Suppose that we are at ancestor $$$u$$$. We will iterate over all the child(say $$$c$$$) of $$$u$$$ such that $$$nax_c \lt i$$$, and add $$$dp[nax_c]$$$ to $$$dp[i]$$$ if $$$nax(u,par_i) \lt nax_c$$$. Suppose $$$track[u][i]$$$ stores the sum of $$$dp[nax_c]$$$ for all $$$c$$$ such that $$$nax_c \lt i$$$. So we should add $$$track[u][i]$$$ to $$$dp[i]$$$. But there is a catch. This way, $$$dp[nax_c]$$$ will also get added $$$dp[i]$$$ even when $$$nax_c \lt nax(u, par_i)$$$. So, we need to subtract some values, which is left as an exercise for the readers. Now, $$$track[u][d] = track[u][d-1] + dp[nax_c]$$$ if $$$nax_c = d$$$. So, instead of keeping a two-dimensional array $$$track$$$, we can just maintain a one-dimensional array $$$track$$$. Note that we will only need the sum of $$$track[u]$$$ for all the ancestors of $$$i$$$, which we can easily calculate using the euler tour.

You can look at the attached code for the implementation details.

The intended time complexity is $$$O(n \cdot \log(n))$$$.

#include <bits/stdc++.h>   
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;   
using namespace std;
#define ll long long
#define pb push_back                  
#define mp make_pair          
#define nline "\n"                            
#define f first                                            
#define s second                                             
#define pll pair<ll,ll> 
#define all(x) x.begin(),x.end()     
const ll MOD=998244353;
const ll MAX=1000100;
struct FenwickTree{
    vector<ll> bit; 
    ll n;
    FenwickTree(ll n){
        this->n = n;
        bit.assign(n, 0);
    }
    FenwickTree(vector<ll> a):FenwickTree(a.size()){  
        ll x=a.size();
        for(size_t i=0;i<x;i++)
            add(i,a[i]);
    }
    ll sum(ll r) {
        ll ret=0;
        for(;r>=0;r=(r&(r+1))-1){
            ret+=bit[r];
            ret%=MOD;
        }
        return ret;
    }
    ll sum(ll l,ll r) {
        if(l>r)
            return 0;
        ll val=sum(r)-sum(l-1)+MOD;
        val%=MOD;
        return val;
    }
    void add(ll idx,ll delta) {
        for(;idx<n;idx=idx|(idx+1)){
            bit[idx]+=delta;
            bit[idx]%=MOD;
        }
    }
};
vector<vector<ll>> adj;
vector<ll> tin(MAX,0),tout(MAX,0);
vector<ll> parent(MAX,0);
vector<ll> overall_max(MAX,0);
ll now=1;
vector<ll> jump_to(MAX,0),sub(MAX,0);
void dfs(ll cur,ll par){
    parent[cur]=par;
    tin[cur]=now++;
    overall_max[cur]=cur;
    for(auto chld:adj[cur]){
        if(chld!=par){
            jump_to[chld]=max(jump_to[cur],cur);
            dfs(chld,cur);
            overall_max[cur]=max(overall_max[cur],overall_max[chld]);
        }
    }
    tout[cur]=now++;
}
vector<ll> dp(MAX,0);
void solve(){ 
    ll n; cin>>n;
    adj.clear();
    adj.resize(n+5);
    for(ll i=1;i<=n-1;i++){
        ll u,v; cin>>u>>v;
        adj[u].push_back(v);  
        adj[v].push_back(u);
    }  
    now=1;      
    dfs(1,0);      
    ll ans=0;    
    FenwickTree enter_time(now+5),exit_time(now+5);
    overall_max[0]=MOD;   
    dp[0]=1;    
    for(ll i=1;i<=n;i++){
        ll p=jump_to[i];
        dp[i]=(enter_time.sum(0,tin[i])-exit_time.sum(0,tin[i])-sub[p]+dp[p]+MOD+MOD)%MOD;
        if(p>i){  
            dp[i]=0;
        }
        ll node=i;
        while(overall_max[node]==i){
            node=parent[node];
        }
        enter_time.add(tin[node],dp[i]);
        exit_time.add(tout[node],dp[i]);
        sub[i]=(enter_time.sum(0,tin[i])-exit_time.sum(0,tin[i])+MOD)%MOD;
    }
    cout<<dp[n]<<nline;
    return;  
}                                       
int main()                                                                               
{       
    ios_base::sync_with_stdio(false);                         
    cin.tie(NULL);                               
    #ifndef ONLINE_JUDGE                 
    freopen("input.txt", "r", stdin);                                           
    freopen("output.txt", "w", stdout);      
    freopen("error.txt", "w", stderr);                        
    #endif     
    ll test_cases=1;                 
    cin>>test_cases;
    while(test_cases--){
        solve();
    }
    cout<<fixed<<setprecision(10);
    cerr<<"Time:"<<1000*((double)clock())/(double)CLOCKS_PER_SEC<<"ms\n"; 
}  

$$$\operatorname{MEX}$$$ of the path from $$$u$$$ to $$$v$$$ will be the minimum value over all the nodes of $$$T$$$ which do not lie on the path from $$$u$$$ to $$$v$$$.

$$$p_1$$$ and $$$p_2$$$ are associated with the Euler tour

$$$p_1$$$ is the permutation of $$$[1,2, \ldots n]$$$ sorted in increasing order on the basis on $$$tin$$$ time observed during Euler tour. Similarly, $$$p_2$$$ is the permutation of $$$[1,2, \ldots n]$$$ sorted in increasing order based on $$$tout$$$ time. Note that any Euler tour is fine.

Now we have $$$p_1$$$ and $$$p_2$$$ with us. Suppose we need to find $$$\operatorname{MEX}$$$ of the path from $$$u$$$ to $$$v$$$. Assume that $$$tin_u \lt tin_v$$$ for convenience. Assume $$$T'$$$ is the forest we get if we remove all the nodes on the path from $$$u$$$ to $$$v$$$ from $$$T$$$. Our goal is to find the minimum value over all the nodes in $$$T'$$$. Assume that $$$T'$$$ is non-empty, as $$$\operatorname{MEX}$$$ will be $$$n$$$ if $$$T'$$$ is empty. Assume $$$LCA$$$ is the lowest common ancestor of $$$u$$$ and $$$v$$$. Suppose $$$m$$$ is the child of $$$LCA(u,v)$$$, which lies on the path from $$$v$$$ to $$$LCA(u,v)$$$.

Let us consider some groups of nodes $$$p$$$ such that.

1. $$$tout_p \lt tin_u$$$
2. $$$tin_u \lt tin_p \lt tin_m$$$
3. $$$tin_m \lt tout_p \lt tin_v$$$
4. $$$tin_v \lt tin_p$$$
5. $$$tout_{LCA} \lt tout_p$$$

Note that we have precisely $$$5$$$ groups, with the $$$i$$$-th group consisting of only those nodes which satisfy the $$$i$$$-th condition.

Here comes the interesting claim. All nodes in $$$T'$$$ are present in atleast one of the above $$$5$$$ groups. There does not exist a node $$$p$$$ such that $$$p$$$ is on the path from $$$u$$$ from $$$v$$$ and $$$p$$$ is present in any of the groups.

Now, it is not hard to observe that if we consider the nodes of any group, they will form a continuous segment in either $$$p_1$$$ or $$$p_2$$$. So we can cover each group in a single query. Hence, we can find the $$$\operatorname{MEX}$$$ of the path in any round using atmost $$$5$$$ queries.

#include <bits/stdc++.h>   
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;   
using namespace std;
#define ll long long
const ll INF_ADD=1e18;
#define pb push_back                  
#define mp make_pair          
#define nline "\n"                            
#define f first                                            
#define s second                                             
#define pll pair<ll,ll> 
#define all(x) x.begin(),x.end()     
const ll MOD=998244353;
const ll MAX=200200;   
vector<ll> adj[MAX];
ll now=0,till=20;
vector<ll> tin(MAX,0),tout(MAX,0),depth(MAX,0);
vector<vector<ll>> jump(MAX,vector<ll>(till+1,0));
void dfs(ll cur,ll par){
    jump[cur][0]=par;
    for(ll i=1;i<=till;i++)
        jump[cur][i]=jump[jump[cur][i-1]][i-1];
    tin[cur]=++now;
    for(ll chld:adj[cur]){
        if(chld!=par){
            depth[chld]=depth[cur]+1;
            dfs(chld,cur);
        }
    }
    tout[cur]=++now;
} 
bool is_ancestor(ll a,ll b){
    if((tin[a]<=tin[b])&&(tout[a]>=tout[b]))
        return 1;
    return 0;
} 
ll lca(ll a,ll b){
    if(is_ancestor(a,b))
        return a;
    for(ll i=till;i>=0;i--){
        if(!is_ancestor(jump[a][i],b))
            a=jump[a][i];
    }
    return jump[a][0];
}  
void solve(){
    ll n; cin>>n;
    ll m; cin>>m;
    vector<ll> a(n+5);
    for(ll i=1;i<=n-1;i++){
        ll u,v; cin>>u>>v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    now=1;
    dfs(1,1);
    vector<ll> p(n),q(n);
    for(ll i=0;i<n;i++){
        p[i]=q[i]=i+1;
    }
    sort(all(p),[&](ll l,ll r){
        return tin[l]<tin[r];
    });
    sort(all(q),[&](ll l,ll r){
        return tout[l]<tout[r];
    });
    for(auto it:p){
        cout<<it<<" ";
    }
    cout<<endl;
    for(auto it:q){
        cout<<it<<" ";
    }
    cout<<endl;
    auto query_p=[&](ll l,ll r){
        ll left_pos=n+1,right_pos=-1;
        for(ll i=0;i<n;i++){
            ll node=p[i];
            if((tin[node]>=l) and (tin[node]<=r)){
                left_pos=min(left_pos,i);
                right_pos=i;
            }
        }
        ll now=n+5;
        if(right_pos!=-1){
            left_pos++,right_pos++;
            cout<<"? 1 "<<left_pos<<" "<<right_pos<<endl;
            cin>>now;
        }
        return now;
    };
    auto query_q=[&](ll l,ll r){
        ll left_pos=n+1,right_pos=-1;
        for(ll i=0;i<n;i++){
            ll node=q[i];
            if((tout[node]>=l) and (tout[node]<=r)){
                left_pos=min(left_pos,i);
                right_pos=i;
            }
        }
        ll now=n+5;
        if(right_pos!=-1){
            left_pos++,right_pos++;
            cout<<"? 2 "<<left_pos<<" "<<right_pos<<endl;
            cin>>now;
        }
        return now;
    };
    while(m--){
        ll u,v; cin>>u>>v;
        if(tout[u]>tout[v]){
            swap(u,v);
        }
        ll lca_node=lca(u,v);
        ll ans=n;
        if(lca_node==v){
            ans=min(ans,query_q(1,tin[u]));
            ans=min(ans,query_p(tin[u]+1,tout[v]));
            ans=min(ans,query_q(tout[v]+1,now));
            cout<<"! "<<ans<<endl;
            ll x; cin>>x;
            continue;
        }
        ans=min(ans,query_q(1,tin[u]));
        ll consider=v;
        for(auto it:adj[lca_node]){
            if(is_ancestor(lca_node,it) and is_ancestor(it,v)){
                consider=it;
            }
        }
        ans=min(ans,query_p(tin[u]+1,tin[consider]-1));
        ans=min(ans,query_q(tin[consider],tin[v]));
        ans=min(ans,query_p(tin[v]+1,tout[lca_node]));
        ans=min(ans,query_q(tout[lca_node]+1,now));
        cout<<"! "<<ans<<endl;
        ll x; cin>>x;
    }
    for(ll i=1;i<=n;i++){
        adj[i].clear();
    }
    return;  
}                                       
int main()                                                                               
{     
    ios_base::sync_with_stdio(false);                         
    cin.tie(NULL); 
    #ifndef ONLINE_JUDGE                 
    freopen("input.txt", "r", stdin);                                           
    freopen("output.txt", "w", stdout);      
    freopen("error.txt", "w", stderr);                        
    #endif  
    ll test_cases=1;                 
    cin>>test_cases;
    while(test_cases--){
        solve();
    }
    cout<<fixed<<setprecision(10);
    cerr<<"Time:"<<1000*((double)clock())/(double)CLOCKS_PER_SEC<<"ms\n"; 
}  

A167510

It is convinient here to assign weights to $$$\mathtt{0} \to -1$$$ and $$$\mathtt{1} \to 1$$$.

Given a string $$$t$$$, we can define the prefix sum $$$p$$$ of it's weights. For example, if $$$t=\mathtt{0010111}$$$, then $$$p=[0,-1,-2,-1,-2,-1,0,1]$$$. So that if $$$t$$$ is bad and $$$i$$$ is a index that violates the definition, then $$$\max(p_0,p_1,\ldots,p_{i-1}) \lt \min(p_i,p_{i+1},\ldots,p_n)$$$ if $$$s_i = \mathtt{0}$$$ or $$$\min(p_0,p_1,\ldots,p_{i-1}) \gt \max(p_i,p_{i+1},\ldots,p_n)$$$ if $$$s_i = \mathtt{1}$$$.

Naturally, it is convinient to assume that $$$p_0 \lt p_n$$$. All string with $$$p_0 = p_n$$$ are clearly good and $$$p_0 \gt p_n$$$ is handled similarly. For strings with $$$p_0 \lt p_n$$$, the condition can only be violated on an index with $$$s_i = \mathtt{0}$$$.

The solution works using PIE. Let us fix a set of positions $$$I$$$ that are bad, so that if $$$i \in I$$$, then $$$s_i =\mathtt{0}$$$ and $$$\max(p_0,p_1,\ldots,p_{i-1}) \lt \min(p_i,p_{i+1},\ldots,p_n)$$$. Then we need to count the number of ways to construct $$$p$$$ satisfying these conditions and then add it to the answer multiplied by $$$(-1)^{|I|}$$$.

Suppose that $$$I = i_1,i_2,\ldots,i_k$$$. $$$t[1,i_1]$$$ and $$$t[i_k,n]$$$ need to be a ballot sequence of length $$$i_1-1$$$ and $$$n-i_k$$$ respectively (A001405, denote it as $$$f(n)$$$) while $$$t[i_j,i_{j+1}]$$$ needs to be bidirectional ballot sequence of length $$$i_{j+1}-i_j-1$$$ (A167510, denote it as $$$g(n)$$$). Note that in our definition of ballot sequence, we are do not require that prefixes and suffixes have strictly more $$$\mathtt{1}$$$ s thatn $$$\mathtt{0}$$$ s. It is the same sequence, but note that we need to shift it by a few places when refering to OEIS.

The first $$$n$$$ terms of $$$f$$$ is easily computed in linear time. We will focus on how to compute the first $$$n$$$ terms of $$$g$$$ in $$$O(n \log^2 n)$$$.

Computing $$$g(n)$$$ Firstly, let us consider the number of bidirectional sequences with $$$\frac{n+k}{2}$$$ $$$\mathtt{1}$$$ s and $$$\frac{n-k}{2}$$$ $$$\mathtt{0}$$$ s. We will imagine this as lattice walks from $$$(0,0)$$$ to $$$(n,k)$$$ where $$$\mathtt{1} \to (1,1)$$$ and $$$\mathtt{0} \to (1,-1)$$$. If we touch the lines $$$y=-1$$$ or $$$y=k+1$$$, the walk is invalid.

We can use the reflection method here, similar to a proof of Catalan. The number of valid walks is $$$#(*) - #(T) + #(TB) - #(TBT) ..... - #(B) + #(BT) - #(BTB) + .....$$$ where $$$#(BTB)$$$ denotes the number of walks that touch the bottom line, then the top line, then the bottom line, and then may continue to touch the top and bottom line after that.

We have $$$#(*) =$$$ $$$\binom{n}{\frac{n+k}{2}}$$$, $$$#(T) =$$$ $$$ \binom{n}{\frac{n+k+2}{2}}$$$, $$$#(TB) =$$$ $$$ \binom{n}{\frac{n+3k+4}{2}}$$$, $$$#(TBT) =$$$ $$$ \binom{n}{\frac{n+3k+6}{2}}$$$, $$$\ldots$$$, $$$#(B) = $$$ $$$\binom{n}{\frac{n+k+2}{2}}$$$, $$$#(BT) =$$$ $$$ \binom{n}{\frac{n+k+4}{2}}$$$, $$$#(BTB) =$$$ $$$ \binom{n}{\frac{n+3k+6}{2}}$$$, $$$\ldots$$$

This already gives us a method to compute $$$g(n)$$$ in $$$O(n \log n)$$$ since for a fixed $$$k$$$, we can compute the above sum in $$$O(\frac{n}{k})$$$, since only the first $$$O(\frac{n}{k})$$$ terms are not $$$0$$$.

First, notice that we can aggregate them sums without iterating on $$$k$$$, for some fixed $$$j$$$, we can find the coefficient $$$c_j$$$ of $$$\binom{n}{\frac{n+j}{2}}$$$ across all $$$k$$$. Notice that this coefficient is independent across all $$$n$$$, so we only need to compute $$$c$$$ once.

Now, note that $$$\binom{n}{\frac{n+z}{2}} = [x^z] (x^{-1} + x) ^ n$$$. So that $$$g(n) = [x^0] C \cdot (x^{-1} + x)^n$$$, where $$$C$$$ is the ogf of $$$c$$$.

From this formulation, we can describe how to compute the first $$$n$$$ terms of $$$g$$$ in $$$O(n \log^2 n)$$$ using Divide and Conquer.

$$$DnC(l,r,V)$$$ computes the $$$g(l) \ldots g(r)$$$ where $$$V$$$ is the coefficents between $$$[l-r,r-l]$$$ of $$$C \cdot (x^{-1} + x)^l$$$. $$$DnC(l,r,V)$$$ will call $$$DnC(l,m,V)$$$ and $$$DnC(m+1,r,V \cdot (x^{-1}+x)^{m-l+1})$$$. We have the reccurence $$$T(n) = 2 T(\frac{n}{2}) + O(n \log n)$$$ so $$$T(n) = O(n \log^2 n)$$$.

Of course, for constant time speedup, you can choose to split the odd and even coefficients, but that is not needed.

It is possible to compute the first $$$n$$$ of $$$g$$$ in $$$O(n \log n)$$$ but it does not improve the overall complexity of the solution.

Final Steps

Now that we obtained the $$$n$$$ terms of $$$f$$$ and $$$g$$$, let us return to the origial problem.

If $$$s_i =\mathtt{0}$$$, define $$$dp_i = f(i-1) - \sum\limits_{s_j = \mathtt{0}} dp_j \cdot g(j-i-1)$$$. Then this contributes $$$f(n-i) \cdot dp_i$$$ to the number of bad strings.

Again, we will use Divide and Conquer to perform this quickly.

Briefly, $$$DnC(l,r)$$$ will compute the values of $$$dp[l,r]$$$ given that contributions from $$$dp[1,l-1]$$$ has been transferred to $$$dp[l,r]$$$ already.

We will call $$$DnC(l,m)$$$, compute the contribution from $$$dp[l,m]$$$ to $$$dp[m+1,r]$$$ using FFT and then call $$$DnC(m+1,r)$$$. The complexity of this is $$$O(n \log^2 n)$$$.

#include <bits/stdc++.h>
using namespace std;

#define int long long
#define ll long long
#define ii pair<int,int>
#define iii tuple<int,int,int>
#define fi first
#define se second
#define endl '\n'
#define debug(x) cout << #x << ": " << x << endl

#define pub push_back
#define pob pop_back
#define puf push_front
#define pof pop_front
#define lb lower_bound
#define ub upper_bound

#define rep(x,start,end) for(int x=(start)-((start)>(end));x!=(end)-((start)>(end));((start)<(end)?x++:x--))
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()

mt19937 rng(chrono::system_clock::now().time_since_epoch().count());


const int MOD=998244353;

ll qexp(ll b,ll p,int m){
    ll res=1;
    while (p){
        if (p&1) res=(res*b)%m;
        b=(b*b)%m;
        p>>=1;
    }
    return res;
}

ll inv(ll i){
	return qexp(i,MOD-2,MOD);
}

ll fix(ll i){
	i%=MOD;
	if (i<0) i+=MOD;
	return i;
}

ll fac[1000005];
ll ifac[1000005];

ll nCk(int i,int j){
	if (i<j) return 0;
	return fac[i]*ifac[j]%MOD*ifac[i-j]%MOD;
}

const ll mod = (119 << 23) + 1, root = 62; // = 998244353
// For p < 2^30 there is also e.g. 5 << 25, 7 << 26, 479 << 21
// and 483 << 21 (same root). The last two are > 10^9.
typedef vector<ll> vl;
void ntt(vl &a) {
	int n = sz(a), L = 31 - __builtin_clz(n);
	static vl rt(2, 1);
	for (static int k = 2, s = 2; k < n; k *= 2, s++) {
		rt.resize(n);
		ll z[] = {1, qexp(root, mod >> s, mod)};
		rep(i,k,2*k) rt[i] = rt[i / 2] * z[i & 1] % mod;
	}
	vector<int> rev(n);
	rep(i,0,n) rev[i] = (rev[i / 2] | (i & 1) << L) / 2;
	rep(i,0,n) if (i < rev[i]) swap(a[i], a[rev[i]]);
	for (int k = 1; k < n; k *= 2)
		for (int i = 0; i < n; i += 2 * k) rep(j,0,k) {
			ll z = rt[j + k] * a[i + j + k] % mod, &ai = a[i + j];
			a[i + j + k] = ai - z + (z > ai ? mod : 0);
			ai += (ai + z >= mod ? z - mod : z);
		}
}
vl conv(const vl &a, const vl &b) {
	if (a.empty() || b.empty()) return {};
	int s = sz(a) + sz(b) - 1, B = 32 - __builtin_clz(s), n = 1 << B;
	int inv = qexp(n, mod - 2, mod);
	vl L(a), R(b), out(n);
	L.resize(n), R.resize(n);
	ntt(L), ntt(R);
	rep(i,0,n) out[-i & (n - 1)] = (ll)L[i] * R[i] % mod * inv % mod;
	ntt(out);
	return {out.begin(), out.begin() + s};
}

int n;
string s;

int c[100005];
int f[100005];

void calc(int l,int r,vector<int> v){
	while (sz(v)>(r-l)*2+1) v.pob();
	
	if (l==r){
		f[l]=v[0];
		return;
	}
	
	int m=l+r>>1;
	
	calc(l,m,vector<int>(v.begin()+(r-m),v.end()));
	vector<int> a;
	int t=m-l+1;
	rep(x,0,t+1) a.pub(nCk(t,x)),a.pub(0);
	
	v=conv(v,a);
	calc(m+1,r,vector<int>(v.begin()+(2*t),v.end()));
}

int ans[100005];

void solve(int l,int r){
	if (l==r) return;
	int m=l+r>>1;
	solve(l,m);
	auto a=conv(vector<int>(ans+l,ans+m+1),vector<int>(f,f+(r-l)));
	rep(x,m+1,r+1) if (s[x]=='0') ans[x]=fix(ans[x]-a[x-1-l]);
	solve(m+1,r);
}


signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin.exceptions(ios::badbit | ios::failbit);
	
	fac[0]=1;
	rep(x,1,1000005) fac[x]=fac[x-1]*x%MOD;
	ifac[1000004]=inv(fac[1000004]);
	rep(x,1000005,1) ifac[x-1]=ifac[x]*x%MOD;
	
	rep(x,1,100005){
		c[x]++;
		
		int curr=x;
		while (curr<100005){
			if (curr+2<100005) c[curr+2]-=2;
			if (curr+4<100005) c[curr+4]++;
			if (curr+2*x+4<100005) c[curr+2*x+4]++;
			curr+=2*x+4;
		}
	}
	
	cin>>n;
	cin>>s;
	
	vector<int> v;
	rep(x,n+1,1) v.pub(fix(c[x]+(x>=2?c[x-2]:0LL)));
	calc(1,n,v);
	f[0]=1;
	
	int fin=qexp(2,n,MOD);
	rep(_,0,2){
		rep(x,0,n) ans[x]=(s[x]=='0')?nCk(x,x/2):0LL;
		solve(0,n-1);
		rep(x,0,n) if (s[x]=='0') fin=fix(fin-ans[x]*nCk((n-x-1),(n-x-1)/2));
		
		for (auto &it:s) it^=1;
	}
	
	cout<<fin<<endl;
}