#include <bits/stdc++.h>
const int MOD = 1e9 + 7;
long long BinPow(int x, int power) {
  if (power == 0) {
    return 1;
  } else if (power == 1) {
    return x % MOD;
  } else {
    long long s = BinPow(x, power / 2);
    if (power % 2) {
      return (s * s % MOD) * x % MOD;
    }
    return s * s % MOD;
  }
}
void Solve() {
  int n = 0, k = 0;
  std::cin >> n >> k;
  std::vector<long long> pow(n + k + 1);
  pow[0] = 1;
  for (int i = 1; i < pow.size(); ++i) {
    pow[i] = pow[i - 1] * 2 % MOD;
  }
  long long ans = 0;
  if (n % 2 == 0) {
    for (int i = 0; i < k; ++i) {
      ans += BinPow(pow[n - 1] - 1, k - 1 - i) * BinPow(pow[n], i) % MOD;
      ans %= MOD;
    }
    ans += BinPow(pow[n - 1] - 1, k);
    ans %= MOD;
  } else {
    ans = BinPow(pow[n - 1] + 1, k);
  }
  std::cout << ans << '\n';
}
int main() {
  int t = 0;
  std::cin >> t;
  while (t--) {
    Solve();
  }
}

### Prompt

Your task is to create 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>
const int MOD = 1e9 + 7;
long long BinPow(int x, int power) {
  if (power == 0) {
    return 1;
  } else if (power == 1) {
    return x % MOD;
  } else {
    long long s = BinPow(x, power / 2);
    if (power % 2) {
      return (s * s % MOD) * x % MOD;
    }
    return s * s % MOD;
  }
}
void Solve() {
  int n = 0, k = 0;
  std::cin >> n >> k;
  std::vector<long long> pow(n + k + 1);
  pow[0] = 1;
  for (int i = 1; i < pow.size(); ++i) {
    pow[i] = pow[i - 1] * 2 % MOD;
  }
  long long ans = 0;
  if (n % 2 == 0) {
    for (int i = 0; i < k; ++i) {
      ans += BinPow(pow[n - 1] - 1, k - 1 - i) * BinPow(pow[n], i) % MOD;
      ans %= MOD;
    }
    ans += BinPow(pow[n - 1] - 1, k);
    ans %= MOD;
  } else {
    ans = BinPow(pow[n - 1] + 1, k);
  }
  std::cout << ans << '\n';
}
int main() {
  int t = 0;
  std::cin >> t;
  while (t--) {
    Solve();
  }
}

```