#include <bits/stdc++.h>
using namespace std;
int f[200010];
void pre_cal() {
  f[0] = 1;
  int i = 1;
  while (i <= 200002) {
    f[i] = f[i - 1] * 2 % 1000000007;
    i++;
  }
}
int power(int a, int b) {
  int res = 1 % 1000000007;
  while (b) {
    if (b & 1) res = (1ll) * res * a % 1000000007;
    a = (1ll) * a * a % 1000000007;
    b >>= 1;
  }
  return res;
}
int main() {
  pre_cal();
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    if (k == 0) {
      cout << 1 << endl;
      continue;
    }
    if (n & 1) {
      cout << power(f[n - 1] + 1, k) << endl;
      continue;
    }
    int ans = power(f[n - 1] - 1, k);
    for (int i = 0; i < k; i++) {
      int res = (long long)power(f[n - 1] - 1, i) * power(f[n], k - i - 1) %
                1000000007;
      ans = (ans + res) % 1000000007;
    }
    cout << ans << endl;
  }
}

### Prompt

Please provide a Cpp coded solution 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;
int f[200010];
void pre_cal() {
  f[0] = 1;
  int i = 1;
  while (i <= 200002) {
    f[i] = f[i - 1] * 2 % 1000000007;
    i++;
  }
}
int power(int a, int b) {
  int res = 1 % 1000000007;
  while (b) {
    if (b & 1) res = (1ll) * res * a % 1000000007;
    a = (1ll) * a * a % 1000000007;
    b >>= 1;
  }
  return res;
}
int main() {
  pre_cal();
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    if (k == 0) {
      cout << 1 << endl;
      continue;
    }
    if (n & 1) {
      cout << power(f[n - 1] + 1, k) << endl;
      continue;
    }
    int ans = power(f[n - 1] - 1, k);
    for (int i = 0; i < k; i++) {
      int res = (long long)power(f[n - 1] - 1, i) * power(f[n], k - i - 1) %
                1000000007;
      ans = (ans + res) % 1000000007;
    }
    cout << ans << endl;
  }
}

```