#include "bits/stdc++.h"
using namespace std;
//#include "atcoder/all"
//using namespace atcoder;
#define int long long
#define REP(i, n) for (int i = 0; i < (int)n; ++i)
#define RREP(i, n) for (int i = (int)n - 1; i >= 0; --i)
#define FOR(i, s, n) for (int i = s; i < (int)n; ++i)
#define RFOR(i, s, n) for (int i = (int)n - 1; i >= s; --i)
#define ALL(a) a.begin(), a.end()
#define IN(a, x, b) (a <= x && x < b)
template<class T>istream&operator >>(istream&is,vector<T>&vec){for(T&x:vec)is>>x;return is;}
template<class T>inline void out(T t){cout << t << "\n";}
template<class T,class... Ts>inline void out(T t,Ts... ts){cout << t << " ";out(ts...);}
template<class T>inline bool CHMIN(T&a,T b){if(a > b){a = b;return true;}return false;}
template<class T>inline bool CHMAX(T&a,T b){if(a < b){a = b;return true;}return false;}
constexpr int INF = 1e18;

// a^b
long long modpow(long long a, long long n, long long mod) {
	long long res = 1;
	while (n > 0) {
		if (n & 1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}

// a^-1
long long modinv(long long a, long long m) {
	long long b = m, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}

// a^x ≡ b (mod. m) となる最小の正の整数 x を求める
long long modlog(long long a, long long b, int m) {
	a %= m, b %= m;

	// calc sqrt{M}
	long long lo = -1, hi = m;
	while (hi - lo > 1) {
		long long mid = (lo + hi) / 2;
		if (mid * mid >= m) hi = mid;
		else lo = mid;
	}
	long long sqrtM = hi;

	// {a^0, a^1, a^2, ..., a^sqrt(m)} 
	map<long long, long long> apow;
	long long amari = 1;
	for (long long r = 0; r < sqrtM; ++r) {
		if (!apow.count(amari)) apow[amari] = r;
		(amari *= a) %= m;
	}

	// check each A^p
	long long A = modpow(modinv(a, m), sqrtM, m);
	amari = b;
	for (long long q = 0; q < sqrtM; ++q) {
		if (apow.count(amari)) {
			long long res = q * sqrtM + apow[amari];
			if (res > 0) return res;
		}
		(amari *= A) %= m;
	}

	// no solutions
	return -1;
}

// 返り値: a と b の最大公約数
// ax + by = gcd(a, b) を満たす (x, y) が格納される
long long extGCD(long long a, long long b, long long &x, long long &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    long long d = extGCD(b, a%b, y, x);
    y -= a/b * x;
    return d;
}

void solve() {
	int X, K;
	cin >> X >> K;
	int MOD = 1000000007;
	int x, y;
	int e = extGCD(MOD - 1, K, x, y);
	out(modpow(X, (y % (MOD - 1) + (MOD - 1)) % (MOD - 1), MOD));
}

signed main(){
	int T;
	cin >> T;
	while(T--) solve();
}