We can simply solve this using loop by Complexity of $$$O(n)$$$ or $$$O(n\log_{}{n})$$$ which must result in TLE because of the constraint over $$$n (1\leq n\leq 10^9)$$$. So we need to find a better solution and here is the Mathematics to find for you a better solution!

We can write $$$x+2\cdot x^2+3\cdot x^3+4\cdot x^4+\ldots+n\cdot x^n$$$ in this way-

$$$x+$$$

$$$2\cdot x^2+$$$

$$$3\cdot x^3+$$$

$$$4\cdot x^4+$$$

$$$\ldots$$$

$$$n\cdot x^n$$$

$$$=$$$

$$$x+$$$

$$$x^2+x^2+$$$

$$$x^3+x^3+x^3+$$$

$$$x^4+x^4+x^4+x^4+$$$

$$$\ldots$$$

$$$x^n+x^n+x^n+x^n+x^n+\ldots+x^n [n$$$ $$$times]$$$

Now if we vertically add columnwise, we get

$$$= (x+x^2+x^3+x^4+\ldots+x^n)+(x^2+x^3+x^4+\ldots+x^n)+(x^3+x^4+\ldots x^n)+(x^4+\ldots x^n)+\ldots +(x^n)$$$

$$$= \sum_{i=1}^{n}{x^i} + \sum_{i=2}^{n}{x^i} + \sum_{i=3}^{n}{x^i} + \sum_{i=4}^{n}{x^i} + \ldots + \sum_{i=n}^{n}{x^i}$$$

$$$= \sum_{j=1}^{n}{\sum_{i=j}^{n}{x^i}}$$$

$$$= \sum_{j=1}^{n}{(\sum_{i=1}^{n}{x^i}-\sum_{i=1}^{j-1}{x^i})}$$$

$$$= \sum_{j=1}^{n}{(\frac{x(x^n-1)}{x-1}-\frac{x(x^{j-1}-1}{x-1})}$$$

$$$= \sum_{j=1}{n}{\frac{x^{n+1}-x^j}{x-1}}$$$

$$$= \sum_{j=1}{n}{\frac{x^{n+1}}{x-1}}-\sum_{j=1}{n}{\frac{x^j}{x-1}}$$$

$$$= n\cdot \frac{x^{n+1}}{x-1} - x\cdot \frac{x^n-1}{(x-1)^2}$$$

$$$= \frac{n\cdot x^{n+1}\cdot(x-1)-x^{n+1}+x}{(x-1)^2}$$$

$$$= \frac{n\cdot x^{n+2}-n\cdot x^{n+1}-x^{n+1}+x}{(x-1)^2}$$$

$$$= \frac{n\cdot x^{n+2}-(n+1)\cdot x^{n+1}+x}{(x-1)^2}$$$

So this is our final mathematical expression which is our answer!!! For modulo, the numerator should be modulo of 998244353 (in non-negative) and the denominator should be inverse modulo of 998244353. Modulo of their product is our answer!

So now, what about complexity?

To determine Powers, we will use binary power function. So complexity ultimately $$$O(\log_{}{n})$$$ !

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

#define ll long long int
#define ul unsigned long long int
#define ld long double
#define TT int TC;cin>>TC;for(int ti=1;ti<=TC;ti++)
#define TT1 int TC=1;for(int ti=1;ti<=TC;ti++)
#define fastio ios_base::sync_with_stdio(0); cin.tie(0);
#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define ans_YN if(ans)YES;else NO;
#define pb push_back
#define pr_q priority_queue
#define F first
#define S second
#define MP make_pair
#define UB upper_bound
#define LB lower_bound
#define B begin()
#define E end()
//#define endl "\n"
ll ceill(ll a,ll b){if(a>=0){if(a%b==0)return a/b;else return a/b+1;}else return a/b;}//Here b>0 for being divisor
ll flr(ll a, ll b){if(a>=0){return a/b;} else {if(a%b==0) return a/b; else return a/b-1;}}//Here b>0 for being divisor
ll GCD(ll a, ll b){ ll ans=__gcd(a,b);if(ans<0)ans=-ans;return ans;}
ll LCM(ll a, ll b){ return (a*b)/(GCD(a,b));}
ll rem(ll a, ll b){ if(a>=0) return a%b; else{a=-a; return (b-a%b)%b;}}
int const mod=998244353;
int const AND=1073741823;

ll POW(ll a, ll b){
    if(b==0) return 1;
    if(b==1) return a;
    ll p=POW(a,b/2)%mod;
    if(!(b&1))return (p*p)%mod;
    else return (((p*p)%mod)*a)%mod;
}

//Modular Multiplicative Inverse
ll inversemod(ll x, ll m){
    // (x^(m-2))%m = x^(-1) mod m
    // Condition GCD(x,m)=1
    ll ans=(POW(x,m-2))%m;
    return ans;
}

int main()
{
    fastio
    TT
    {
        ll x,n;
        cin>>x>>n;
        //x+2x^2+...+nx^n
        ll ans;
        if(x==1)ans=((n%mod*(n+1)%mod)%mod*inversemod(2LL,mod))%mod;
        else ans=(rem(((n*POW(x,n+2))%mod-((n+1)*POW(x,n+1))%mod+x%mod),mod)*inversemod(POW(x-1,2),mod))%mod;
        cout<<ans<<endl;
    }
    return 0;
}