#include <bits/stdc++.h>
using namespace std;
vector<long long> fact((long long)3e5);
vector<long long> ifact((long long)3e5);
long long modexp(long long x, long long y) {
  x %= (long long)1e9 + 7;
  if (x == 0) return 0;
  long long res = 1;
  while (y) {
    if (y & 1) {
      res = (x * res);
      res %= (long long)1e9 + 7;
    }
    y /= 2;
    x = (x * x);
    x %= (long long)1e9 + 7;
  }
  return res;
}
long long mod_inverse(long long n) { return modexp(n, (long long)1e9 + 7 - 2); }
long long ncr(long long n, long long r) {
  long long ans = (fact[n] * ifact[r]);
  ans %= (long long)1e9 + 7;
  ans = (ans * ifact[n - r]);
  ans %= (long long)1e9 + 7;
  return ans;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  if (k == 0) {
    cout << 1 << endl;
    return;
  }
  long long ans = 2;
  for (long long i = 2; i < n; i += 2) {
    ans += ncr(n, i);
  }
  if (n % 2) {
    ans = modexp(ans, k);
    ans %= (long long)1e9 + 7;
    cout << ans << endl;
    return;
  }
  long long total = modexp(ans - 1, k);
  for (long long i = 1; i <= k; i++) {
    long long curr = modexp(ans - 1, i - 1) * modexp(2, n * (k - i));
    curr %= (long long)1e9 + 7;
    total += curr;
    total %= (long long)1e9 + 7;
  }
  cout << total << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fact[0] = 1;
  for (long long i = 1; i < (long long)3e5; i++) {
    fact[i] = i * fact[i - 1];
    fact[i] %= (long long)1e9 + 7;
    ifact[i] = mod_inverse(fact[i]);
    ifact[i] %= (long long)1e9 + 7;
  }
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

In Cpp, your task is to solve the following problem:

Moamen and Ezzat are playing a game. They create an array a of n non-negative integers where every element is less than 2^k.

Moamen wins if a_1  \&  a_2  \&  a_3  \&  …  \&  a_n ≥ a_1 ⊕ a_2 ⊕ a_3 ⊕ … ⊕ a_n.

Here \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND), and ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

Please calculate the number of winning for Moamen arrays a.

As the result may be very large, print the value modulo 1 000 000 007 (10^9 + 7).

Input

The first line contains a single integer t (1 ≤ t ≤ 5)— the number of test cases. 

Each test case consists of one line containing two integers n and k (1 ≤ n≤ 2⋅ 10^5, 0 ≤ k ≤ 2⋅ 10^5).

Output

For each test case, print a single value — the number of different arrays that Moamen wins with.

Print the result modulo 1 000 000 007 (10^9 + 7).

Example

Input


3
3 1
2 1
4 0


Output


5
2
1

Note

In the first example, n = 3, k = 1. As a result, all the possible arrays are [0,0,0], [0,0,1], [0,1,0], [1,0,0], [1,1,0], [0,1,1], [1,0,1], and [1,1,1].

Moamen wins in only 5 of them: [0,0,0], [1,1,0], [0,1,1], [1,0,1], and [1,1,1].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<long long> fact((long long)3e5);
vector<long long> ifact((long long)3e5);
long long modexp(long long x, long long y) {
  x %= (long long)1e9 + 7;
  if (x == 0) return 0;
  long long res = 1;
  while (y) {
    if (y & 1) {
      res = (x * res);
      res %= (long long)1e9 + 7;
    }
    y /= 2;
    x = (x * x);
    x %= (long long)1e9 + 7;
  }
  return res;
}
long long mod_inverse(long long n) { return modexp(n, (long long)1e9 + 7 - 2); }
long long ncr(long long n, long long r) {
  long long ans = (fact[n] * ifact[r]);
  ans %= (long long)1e9 + 7;
  ans = (ans * ifact[n - r]);
  ans %= (long long)1e9 + 7;
  return ans;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  if (k == 0) {
    cout << 1 << endl;
    return;
  }
  long long ans = 2;
  for (long long i = 2; i < n; i += 2) {
    ans += ncr(n, i);
  }
  if (n % 2) {
    ans = modexp(ans, k);
    ans %= (long long)1e9 + 7;
    cout << ans << endl;
    return;
  }
  long long total = modexp(ans - 1, k);
  for (long long i = 1; i <= k; i++) {
    long long curr = modexp(ans - 1, i - 1) * modexp(2, n * (k - i));
    curr %= (long long)1e9 + 7;
    total += curr;
    total %= (long long)1e9 + 7;
  }
  cout << total << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fact[0] = 1;
  for (long long i = 1; i < (long long)3e5; i++) {
    fact[i] = i * fact[i - 1];
    fact[i] %= (long long)1e9 + 7;
    ifact[i] = mod_inverse(fact[i]);
    ifact[i] %= (long long)1e9 + 7;
  }
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```