#include <bits/stdc++.h>
using namespace std;
int n, k;
long long dp[200005];
long long getmod(long long x) {
  long long tmp = 2;
  long long res = 1;
  while (x) {
    if (x & 1) res = (res * tmp) % 1000000007;
    x >>= 1;
    tmp = (tmp * tmp) % 1000000007;
  }
  return res;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    dp[0] = 1;
    for (int i = 1; i <= k; i++) {
      if (n & 1) {
        dp[i] = (dp[i - 1] * getmod(n - 1) + dp[i - 1]) % 1000000007;
      } else {
        dp[i] =
            (dp[i - 1] * getmod(n - 1) - dp[i - 1] + 1000000007) % 1000000007 +
            getmod((long long)n * (i - 1));
        dp[i] %= 1000000007;
      }
    }
    cout << dp[k] << endl;
  }
  return 0;
}

### 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 n, k;
long long dp[200005];
long long getmod(long long x) {
  long long tmp = 2;
  long long res = 1;
  while (x) {
    if (x & 1) res = (res * tmp) % 1000000007;
    x >>= 1;
    tmp = (tmp * tmp) % 1000000007;
  }
  return res;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    dp[0] = 1;
    for (int i = 1; i <= k; i++) {
      if (n & 1) {
        dp[i] = (dp[i - 1] * getmod(n - 1) + dp[i - 1]) % 1000000007;
      } else {
        dp[i] =
            (dp[i - 1] * getmod(n - 1) - dp[i - 1] + 1000000007) % 1000000007 +
            getmod((long long)n * (i - 1));
        dp[i] %= 1000000007;
      }
    }
    cout << dp[k] << endl;
  }
  return 0;
}

```