#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
long long pow_mod(long long a, long long b) {
  long long res = 1L;
  a = a % MOD;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % MOD;
    }
    b /= 2;
    a = (a * a) % MOD;
  }
  return res;
}
int main() {
  int t, n, k;
  long long ans, a, b, num, inverse, common_power;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    if (n & 1) {
      ans = pow_mod(2, n - 1);
      ans = (ans + 1) % MOD;
      ans = pow_mod(ans, k);
    } else {
      common_power = pow_mod(2, n - 1);
      a = pow_mod(common_power * 2, k);
      b = pow_mod(common_power - 1, k);
      num = (a - b);
      if (num < 0) num += MOD;
      inverse = pow_mod(common_power + 1, 1000000005);
      ans = (inverse * num) % MOD;
      ans = (ans + b) % MOD;
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
long long pow_mod(long long a, long long b) {
  long long res = 1L;
  a = a % MOD;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % MOD;
    }
    b /= 2;
    a = (a * a) % MOD;
  }
  return res;
}
int main() {
  int t, n, k;
  long long ans, a, b, num, inverse, common_power;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    if (n & 1) {
      ans = pow_mod(2, n - 1);
      ans = (ans + 1) % MOD;
      ans = pow_mod(ans, k);
    } else {
      common_power = pow_mod(2, n - 1);
      a = pow_mod(common_power * 2, k);
      b = pow_mod(common_power - 1, k);
      num = (a - b);
      if (num < 0) num += MOD;
      inverse = pow_mod(common_power + 1, 1000000005);
      ans = (inverse * num) % MOD;
      ans = (ans + b) % MOD;
    }
    cout << ans << "\n";
  }
  return 0;
}

```