#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1e9 + 7;
long long int binpow(long long int a, long long int b) {
  long long int r = 1;
  while (b) {
    if (b & 1) r = (r % MOD * a % MOD) % MOD;
    a = (a % MOD * a % MOD) % MOD;
    b /= 2;
  }
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1, testcases = 1;
  cin >> t;
  while (t >= testcases) {
    long long int n, k;
    cin >> n >> k;
    long long int ans = 0;
    if (n % 2) {
      ans = (binpow(2, n - 1) + 1) % MOD;
      ans = binpow(ans, k);
    } else {
      long long int dp[k + 1];
      dp[0] = 1;
      long long int c = (binpow(2, n - 1) - 1 + MOD) % MOD;
      for (long long int i = 1; i <= k; i++) {
        long long int p = n * (i - 1);
        long long int o = binpow(2, p);
        dp[i] = (o % MOD + (c % MOD * dp[i - 1] % MOD) % MOD) % MOD;
      }
      ans = dp[k];
    }
    cout << ans << "\n";
    testcases++;
  }
  cerr << "Time Taken:" << (float)clock() / CLOCKS_PER_SEC * 1000 << "ms"
       << "\n";
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
using namespace std;
const long long int MOD = 1e9 + 7;
long long int binpow(long long int a, long long int b) {
  long long int r = 1;
  while (b) {
    if (b & 1) r = (r % MOD * a % MOD) % MOD;
    a = (a % MOD * a % MOD) % MOD;
    b /= 2;
  }
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1, testcases = 1;
  cin >> t;
  while (t >= testcases) {
    long long int n, k;
    cin >> n >> k;
    long long int ans = 0;
    if (n % 2) {
      ans = (binpow(2, n - 1) + 1) % MOD;
      ans = binpow(ans, k);
    } else {
      long long int dp[k + 1];
      dp[0] = 1;
      long long int c = (binpow(2, n - 1) - 1 + MOD) % MOD;
      for (long long int i = 1; i <= k; i++) {
        long long int p = n * (i - 1);
        long long int o = binpow(2, p);
        dp[i] = (o % MOD + (c % MOD * dp[i - 1] % MOD) % MOD) % MOD;
      }
      ans = dp[k];
    }
    cout << ans << "\n";
    testcases++;
  }
  cerr << "Time Taken:" << (float)clock() / CLOCKS_PER_SEC * 1000 << "ms"
       << "\n";
  return 0;
}

```