#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, P = 1e9 + 7;
int T, n, k;
int dp[N];
int power(int a, int b, int c = 1) {
  for (; b; b >>= 1, a = 1ll * a * a % P)
    if (b & 1) c = 1ll * c * a % P;
  return c;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> T;
  while (T--) {
    cin >> n >> k;
    memset(dp, 0, sizeof(dp));
    dp[0] = 1;
    int prod = 1, pw1 = power(2, n), pw2 = power(2, n - 1);
    for (int i = 1; i <= k; i++) {
      if (n % 2 == 0) {
        dp[i] = 1ll * dp[i - 1] * (pw2 - 1 + P) % P;
        dp[i] = (dp[i] + prod) % P;
      } else {
        dp[i] = 1ll * dp[i - 1] * (pw2 + 1) % P;
      }
      prod = 1ll * prod * pw1 % P;
    }
    cout << dp[k] << endl;
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to 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;
const int N = 2e5 + 10, P = 1e9 + 7;
int T, n, k;
int dp[N];
int power(int a, int b, int c = 1) {
  for (; b; b >>= 1, a = 1ll * a * a % P)
    if (b & 1) c = 1ll * c * a % P;
  return c;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> T;
  while (T--) {
    cin >> n >> k;
    memset(dp, 0, sizeof(dp));
    dp[0] = 1;
    int prod = 1, pw1 = power(2, n), pw2 = power(2, n - 1);
    for (int i = 1; i <= k; i++) {
      if (n % 2 == 0) {
        dp[i] = 1ll * dp[i - 1] * (pw2 - 1 + P) % P;
        dp[i] = (dp[i] + prod) % P;
      } else {
        dp[i] = 1ll * dp[i - 1] * (pw2 + 1) % P;
      }
      prod = 1ll * prod * pw1 % P;
    }
    cout << dp[k] << endl;
  }
  return 0;
}

```