Lets see when the answer is $$$0$$$. If two consecutive non missing elements are not following the condition i.e. $$$a_{i + 1} \leq a_{i}$$$. If two elements one to the left of a missing position, lets call is $$$a$$$ and one to the right, lets call it $$$b$$$ are not following the condition i.e. $$$b \leq a + 1$$$.

In any other scenario, the answer always exists. We can place $$$b - a - 1$$$ elements at each missing position ($$$a$$$ and $$$b$$$ as defined above) and multiply all these values to get the answer.

#include <bits/stdc++.h>
using namespace std;
#define int long long
typedef unsigned long long ull ;

void solve()
{ 
    int n  ;
    cin >> n  ; 

   bitset<65> bt1(n); 
 
   set<int> possible ; 
   int previ= 0 ; 
   for(int i =0 ;i<=64; i++ ) 
   {
       if( bt1[i] ==  0 )
       {
       for(int j = i-1 ; j>=previ  ;j-- ) 
       {
          bt1[j] = 0 ; 
       }
       ull t2= bt1.to_ullong() ; 
       possible.insert(t2); 
       previ = i; 
       }
   }

   cout << possible.size()  << "\n"; 
   for( auto x:  possible) 
   {
      cout << x << " " ; 
   }
   cout << "\n" ;

}


signed main(){
       ios_base::sync_with_stdio(false);
       cin.tie(NULL);
       
       int tt;
       cin >> tt;
       while(tt--)
        solve();
    return 0;
}

If a bit is unset it remains unset.

Notice when a new AND value is added to the set.

The AND Sequence, though infinite, involves a decreasing operation. The solution employs a greedy approach: if a continuous segment of ones is set in the binary representation of the last number added to the set, they all become unset simultaneously.Thus, our approach is simple: set all consecutive segments of ones from right to left to $$$0$$$, and each time, add the resulting number to the set. Given that $$$n \le 10^{18} $$$, we have at most $$$\mathcal{O}(60)$$$ operations per test case, as there are a maximum of $$$60$$$ bits.Be careful that are sequence start from $$$n$$$&($$$n$$$+$$$1$$$).

#include <bits/stdc++.h>
using namespace std;
#define int long long
typedef unsigned long long ull ;

void solve()
{ 
    int n  ;
    cin >> n  ; 

   bitset<65> bt1(n); 
 
   set<int> possible ; 
   int previ= 0 ; 
   for(int i =0 ;i<=64; i++ ) 
   {
       if( bt1[i] ==  0 )
       {
       for(int j = i-1 ; j>=previ  ;j-- ) 
       {
          bt1[j] = 0 ; 
       }
       ull t2= bt1.to_ullong() ; 
       possible.insert(t2); 
       previ = i; 
       }
   }

   cout << possible.size()  << "\n"; 
   for( auto x:  possible) 
   {
      cout << x << " " ; 
   }
   cout << "\n" ;

}


signed main(){
       ios_base::sync_with_stdio(false);
       cin.tie(NULL);
       
       int tt;
       cin >> tt;
       while(tt--)
        solve();
    return 0;
}

Lets rephrase the problem first. If we can colour a permutation in $$$k$$$ operations, we can always colour it in exactly $$$k$$$ operations. So, I will try to colour permutations in least number of operations possible. I will always select a maximal increasing subarray at any instance and colour it entirely. Now, the problem reduces down to finding minimum number of increasing subarrays I can split the permutation into, lets call it $$$x$$$ which can be coloured in $$$x$$$ operations.

We can see that if a permutation can be split into $$$x$$$ increasing subarrays. It has $$$x - 1$$$ number of adjacent value pairs where $$$a_i \gt a_{i + 1}$$$. Lets call these kinds of adjacent value pairs as "bad" pairs. So, we will try to find the number of permutations for each value of $$$k$$$ from $$$0$$$ to $$$n - 1$$$ where $$$k$$$ denotes the number of bad pairs.

We will use dynamic programming to do this. $$$dp(i)(k)$$$ stores the number of permutations of length $$$i$$$ with $$$k$$$ bad pairs. Now, if we want to place the element $$$i + 1$$$ to make permutations of length $$$i + 1$$$. I can place it as the last element of any $$$k + 1$$$ increasing subarrays to retain the number of bad pairs or place it anywhere else to increase the number of bad pairs by $$$1$$$.

#include <bits/stdc++.h>
using namespace std;
#define int long long
typedef unsigned long long ull ;

void solve()
{ 
    int n  ;
    cin >> n  ; 

   bitset<65> bt1(n); 
 
   set<int> possible ; 
   int previ= 0 ; 
   for(int i =0 ;i<=64; i++ ) 
   {
       if( bt1[i] ==  0 )
       {
       for(int j = i-1 ; j>=previ  ;j-- ) 
       {
          bt1[j] = 0 ; 
       }
       ull t2= bt1.to_ullong() ; 
       possible.insert(t2); 
       previ = i; 
       }
   }

   cout << possible.size()  << "\n"; 
   for( auto x:  possible) 
   {
      cout << x << " " ; 
   }
   cout << "\n" ;

}


signed main(){
       ios_base::sync_with_stdio(false);
       cin.tie(NULL);
       
       int tt;
       cin >> tt;
       while(tt--)
        solve();
    return 0;
}