description: 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].
solution: #include <bits/stdc++.h>
  long long power(long long a, int k) {
    long long p = 1;
    while (k) {
      if (k & 1) p = p * a % 1000000007;
      a = a * a % 1000000007;
      k >>= 1;
    }
    return p;
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
      int n, k, i, ans;
      scanf("%d%d", &n, &k);
      if (k == 0)
        printf("1\n");
      else if (n % 2 == 1)
        printf("%lld\n", power(power(2, n - 1) + 1, k));
      else {
        ans = power(power(2, n - 1) - 1, k);
        for (i = 0; i < k; i++)
          ans = (ans +
                 power(power(2, n - 1) - 1, i) * power(power(2, n), k - 1 - i)) %
                1000000007;
        printf("%d\n", ans);
      }
    }
    return 0;
  }
  