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

```