#include <bits/stdc++.h>
long long luth[200005];
class game {
 private:
  int n, k;

 public:
  long long dp[200005][5];
  long long res = 0;
  void in() {
    int x, y;
    std::cin >> x >> y;
    n = x;
    k = y;
  }
  void solve() {
    dp[0][0] = luth[n];
    for (int i = 1; i <= k; i++) {
      dp[i][0] = (dp[i - 1][0] * luth[n]) % 1000000007;
    }
    if (k == 0) {
      res = 1;
      return;
    }
    if (k == 1) {
      if (n % 2 == 1) {
        res = ((luth[n - 1]) + 1) % 1000000007;
      } else {
        res = (luth[n - 1]) % 1000000007;
      }
      return;
    }
    if (n % 2 == 1) {
      dp[0][1] = (luth[n - 1] + 1) % 1000000007;
      for (int i = 1; i < k; i++) {
        dp[i][1] = (dp[i - 1][1] * (luth[n - 1] + 1)) % 1000000007;
      }
    } else {
      dp[0][1] = (luth[n - 1]) % 1000000007;
      for (int i = 1; i < k; i++) {
        dp[i][1] =
            ((dp[i - 1][1] * (luth[n - 1] - 1)) + dp[i - 1][0]) % 1000000007;
      }
    }
    res = dp[k - 1][1];
  }
  void out() { std::cout << res << std::endl; }
};
int main() {
  int tests;
  std::cin >> tests;
  luth[0] = 1;
  for (int i = 1; i <= 200003; i++) {
    luth[i] = luth[i - 1] * 2;
    if (luth[i] >= 1000000007) luth[i] = luth[i] % 1000000007;
  }
  for (int i = 0; i < tests; i++) {
    game *obj = new game;
    obj->in();
    obj->solve();
    obj->out();
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 luth[200005];
class game {
 private:
  int n, k;

 public:
  long long dp[200005][5];
  long long res = 0;
  void in() {
    int x, y;
    std::cin >> x >> y;
    n = x;
    k = y;
  }
  void solve() {
    dp[0][0] = luth[n];
    for (int i = 1; i <= k; i++) {
      dp[i][0] = (dp[i - 1][0] * luth[n]) % 1000000007;
    }
    if (k == 0) {
      res = 1;
      return;
    }
    if (k == 1) {
      if (n % 2 == 1) {
        res = ((luth[n - 1]) + 1) % 1000000007;
      } else {
        res = (luth[n - 1]) % 1000000007;
      }
      return;
    }
    if (n % 2 == 1) {
      dp[0][1] = (luth[n - 1] + 1) % 1000000007;
      for (int i = 1; i < k; i++) {
        dp[i][1] = (dp[i - 1][1] * (luth[n - 1] + 1)) % 1000000007;
      }
    } else {
      dp[0][1] = (luth[n - 1]) % 1000000007;
      for (int i = 1; i < k; i++) {
        dp[i][1] =
            ((dp[i - 1][1] * (luth[n - 1] - 1)) + dp[i - 1][0]) % 1000000007;
      }
    }
    res = dp[k - 1][1];
  }
  void out() { std::cout << res << std::endl; }
};
int main() {
  int tests;
  std::cin >> tests;
  luth[0] = 1;
  for (int i = 1; i <= 200003; i++) {
    luth[i] = luth[i - 1] * 2;
    if (luth[i] >= 1000000007) luth[i] = luth[i] % 1000000007;
  }
  for (int i = 0; i < tests; i++) {
    game *obj = new game;
    obj->in();
    obj->solve();
    obj->out();
  }
  return 0;
}

```