#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("unroll-loops")
//#pragma warning(disable : 4996)

//#define ATCODER
#ifdef ATCODER
#include<atcoder/all>
using namespace atcoder;
#endif

#include <algorithm>
#include <utility>
#include <vector>

#ifdef _MSC_VER
#include <intrin.h>
#define __builtin_popcount __popcnt
#define __builtin_popcountll __popcnt64

#endif

#include <limits.h>
#include <math.h>
#include <time.h>

#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <complex>
#include <cstdio>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>

using namespace std;

#define REP(i, n) for (int i = 0; i < (n); ++i)
#define REPR(i, n) for (int i = n - 1; i >= 0; --i)
#define FOR(i, m, n) for (int i = m; i < n; ++i)
#define FORR(i, m, n) for (int i = m - 1; i >= n; --i)
#define SORT(v, n) sort(v, v + n);
#define VSORT(v) sort(v.begin(), v.end());
#define REVERSE(v, n) reverse(v, v + n);
#define VREVERSE(v) reverse(v.begin(), v.end())
#define ll long long
#define print(x) cout << (x) << '\n'
#define pe(x) cout << (x) << " "
#define DEBUG(x) cout << #x << ": " << x << endl
#define lb(v, n) lower_bound(v.begin(), v.end(), (n))
#define ub(v, n) upper_bound(v.begin(), v.end(), (n))
#define int long long
//#define double long double
#define all(x) (x).begin(), (x).end()
#define print_space(v) REP(i, v.size()) cout << v[i] << " \n"[i==(int)v.size()-1]
template <typename T1, typename T2> inline void chmin(T1 & a, T2 b) { if (a > b) a = b; }
template <typename T1, typename T2> inline void chmax(T1& a, T2 b) { if (a < b) a = b; }
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
std::random_device rd;
std::mt19937 mt(rd());
constexpr ll MOD = 998244353;
constexpr int MAX = 440000;
const double pi = acos(-1);
constexpr double EPS = 1e-8;
constexpr ll LINF = 1e17 + 1;
constexpr int INF = 1e9 + 1;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}

// 二項係数計算
long long COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
ll add(ll x, ll y) {
	x += y;
	if (x >= MOD) return x - MOD;
	return x;
}
ll sub(ll x, ll y) {
	x -= y;
	if (x < 0) return x + MOD;
	return x;
}
ll mult(ll x, ll y) {
	return (x * y) % MOD;
}
ll bin_pow(ll x, ll p) {
	if (p == 0) return 1;
	if (p & 1) return mult(x, bin_pow(x, p - 1));
	return bin_pow(mult(x, x), p / 2);
}

//const int mod = 1000000007;
const int mod = 998244353;
struct mint {
	ll x; // typedef long long ll;
	mint(ll x = 0) :x((x%mod + mod) % mod) {}
	mint operator-() const { return mint(-x); }
	mint& operator+=(const mint a) {
		if ((x += a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint a) {
		if ((x += mod - a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this; }
	mint operator+(const mint a) const { return mint(*this) += a; }
	mint operator-(const mint a) const { return mint(*this) -= a; }
	mint operator*(const mint a) const { return mint(*this) *= a; }
	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime mod
	mint inv() const { return pow(mod - 2); }
	mint& operator/=(const mint a) { return *this *= a.inv(); }
	mint operator/(const mint a) const { return mint(*this) /= a; }
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }


void solve() {
	mint t = 2;
	int N, K; cin >> N >> K;
	mint ans = (t.pow(K*N)-t.pow(K*(N - 1)))*N;
	print(ans);
}




signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int q;
	cin >> q;
	while (q--)
	solve();

}