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!

Also, think about the corner case! This function gives a $$$\frac{0}{0}$$$ form for $$$x=1$$$. So it will not work for $$$x=1$$$. We can simply find that for $$$x=1$$$, the series becomes $$$1+2+3+\ldots+n$$$ which equals to $$$\frac{n\cdot(n+1)}{2}$$$.

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;

//Binary Power Function
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;
        //Here rem function used to convert negative mod to non-negative
        cout<<ans<<endl;
    }
    return 0;
}