#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int i, j, k, n, m, x, y, t;
int mi(int x, int y) {
  if (y == 0) return 1;
  int t = mi(x, y >> 1);
  ;
  t = (long long)t * t % mod;
  if (y & 1) t = (long long)t * x % mod;
  return t;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &m);
    int ans = 0;
    x = 1;
    for (i = 0; i < n; i += 2) {
      ans = (ans + x) % mod;
      x = (long long)x * (n - i) % mod * (n - i - 1) % mod *
          mi(i + 1, mod - 2) % mod * mi(i + 2, mod - 2) % mod;
    }
    if (n & 1) {
      ans++;
      printf("%d\n", mi(ans, m));
    } else {
      x = ans;
      ans = mi(ans, m);
      for (i = m; i >= 1; i--)
        ans =
            (ans + (long long)mi(mi(2, n), i - 1) * mi(x, (m - i)) % mod) % mod;
      printf("%d\n", ans);
    }
  }
  return 0;
}

### Prompt

Generate 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 mod = 1000000007;
int i, j, k, n, m, x, y, t;
int mi(int x, int y) {
  if (y == 0) return 1;
  int t = mi(x, y >> 1);
  ;
  t = (long long)t * t % mod;
  if (y & 1) t = (long long)t * x % mod;
  return t;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &m);
    int ans = 0;
    x = 1;
    for (i = 0; i < n; i += 2) {
      ans = (ans + x) % mod;
      x = (long long)x * (n - i) % mod * (n - i - 1) % mod *
          mi(i + 1, mod - 2) % mod * mi(i + 2, mod - 2) % mod;
    }
    if (n & 1) {
      ans++;
      printf("%d\n", mi(ans, m));
    } else {
      x = ans;
      ans = mi(ans, m);
      for (i = m; i >= 1; i--)
        ans =
            (ans + (long long)mi(mi(2, n), i - 1) * mi(x, (m - i)) % mod) % mod;
      printf("%d\n", ans);
    }
  }
  return 0;
}

```