#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const ll modc=998244353;
class mint {
    ll x;
public:
    mint(ll x=0) : x((x%modc+modc)%modc) {}
    mint operator-() const {
      return mint(-x);
    }
    mint& operator+=(const mint& a) {
        if ((x += a.x) >= modc) x -= modc;
        return *this;
    }
    mint& operator-=(const mint& a) {
        if ((x += modc-a.x) >= modc) x -= modc;
        return *this;
    }
    mint& operator*=(const  mint& a) {
        (x *= a.x) %= modc;
        return *this;
    }
    mint operator+(const mint& a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint& a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint& a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    mint inv() const {
        return pow(modc-2);
    }
    mint& operator/=(const mint& a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint& a) const {
        mint res(*this);
        return res/=a;
    }
    bool operator == (const mint& a) const{
        return x == a.x;
    }
    friend ostream& operator<<(ostream& os, const mint& m){
        os << m.x;
        return os;
    }
    friend istream& operator>>(istream& ip, mint &m) {
        ll t;
        ip >> t;
        m = mint(t);
        return ip;
    }
    ll val(){
        return x;
    }
};
void solve(){

    ll N, M;
    cin >> M >> N;
    mint ans;

    //dp(i, j): i行目にj個の宝を埋める方法の総数
    //前の行に2個宝がある(i, j)とき、(1,2),(2,3),(3,4)...(M, 1)のうち、i, jを含む4つが選べない
    vector dp(N+1, vector<mint>(3));
    dp[1][0] = 1;
    dp[1][1] = M;
    dp[1][2] = (M == 2 ? 0LL : M*(M-1)/2-M);
    for (int j=2; j<=N; j++){
        for (int k=0; k<=2; k++){
            for (int l=0; l<=2; l++){
                if (l == 0) dp[j][l] += dp[j-1][k];
                else if (l == 1) dp[j][l] += dp[j-1][k] * (M-k);
                else dp[j][l] += (M == 2 ? 0LL : dp[j-1][k] * ((M-k)*(M-k-1)/2-(M-k*2)));
            }
        }
    }
    ans = dp[N][0] + dp[N][1] + dp[N][2];
    cout << ans << endl;
}

int main(){
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);

    int T;
    cin >> T;
    while(T--) solve();

    return 0;
}