#include <bits/stdc++.h>
long long power(long long a, int k) {
  long long p = 1;
  while (k) {
    if (k & 1) p = p * a % 1000000007;
    a = a * a % 1000000007;
    k >>= 1;
  }
  return p;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, k, i, ans;
    scanf("%d%d", &n, &k);
    if (k == 0)
      printf("1\n");
    else if (n % 2 == 1)
      printf("%lld\n", power(power(2, n - 1) + 1, k));
    else {
      ans = power(power(2, n - 1) - 1, k);
      for (i = 0; i < k; i++)
        ans = (ans +
               power(power(2, n - 1) - 1, i) * power(power(2, n), k - 1 - i)) %
              1000000007;
      printf("%d\n", ans);
    }
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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>
long long power(long long a, int k) {
  long long p = 1;
  while (k) {
    if (k & 1) p = p * a % 1000000007;
    a = a * a % 1000000007;
    k >>= 1;
  }
  return p;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, k, i, ans;
    scanf("%d%d", &n, &k);
    if (k == 0)
      printf("1\n");
    else if (n % 2 == 1)
      printf("%lld\n", power(power(2, n - 1) + 1, k));
    else {
      ans = power(power(2, n - 1) - 1, k);
      for (i = 0; i < k; i++)
        ans = (ans +
               power(power(2, n - 1) - 1, i) * power(power(2, n), k - 1 - i)) %
              1000000007;
      printf("%d\n", ans);
    }
  }
  return 0;
}

```