#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  long long res = 1;
  a = a % 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
      b--;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long fermat_inv(long long y) { return power(y, 1000000007 - 2ll); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long dp[k + 1];
    memset(dp, 0, sizeof(dp));
    dp[0] = 1ll;
    long long ans = 0ll;
    for (int i = 0; i < k; i++) {
      if (n % 2) {
        dp[i + 1] = (dp[i] * power(2ll, n - 1ll)) % 1000000007;
        dp[i + 1] = (dp[i + 1] + dp[i]) % 1000000007;
      } else {
        dp[i + 1] =
            (dp[i] * ((power(2ll, n - 1ll) - 1ll + 1000000007) % 1000000007)) %
            1000000007;
        ans = (ans + (dp[i] * power(power(2ll, k - i - 1), n)) % 1000000007) %
              1000000007;
      }
    }
    ans = (ans + dp[k]) % 1000000007;
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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>
using namespace std;
long long power(long long a, long long b) {
  long long res = 1;
  a = a % 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
      b--;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long fermat_inv(long long y) { return power(y, 1000000007 - 2ll); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long dp[k + 1];
    memset(dp, 0, sizeof(dp));
    dp[0] = 1ll;
    long long ans = 0ll;
    for (int i = 0; i < k; i++) {
      if (n % 2) {
        dp[i + 1] = (dp[i] * power(2ll, n - 1ll)) % 1000000007;
        dp[i + 1] = (dp[i + 1] + dp[i]) % 1000000007;
      } else {
        dp[i + 1] =
            (dp[i] * ((power(2ll, n - 1ll) - 1ll + 1000000007) % 1000000007)) %
            1000000007;
        ans = (ans + (dp[i] * power(power(2ll, k - i - 1), n)) % 1000000007) %
              1000000007;
      }
    }
    ans = (ans + dp[k]) % 1000000007;
    cout << ans << "\n";
  }
  return 0;
}

```