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 n, k;
  long long fastPower(long long base, long long power, long long mod) {
    base = base % mod;
    long long ans = 1;
    while (power) {
      if (power & 1) {
        ans = ans * base;
        ans = ans % mod;
      }
      base = base * base;
      base = base % mod;
      power = power >> 1;
    }
    ans = ans % mod;
    return ans;
  }
  int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
      scanf("%lld %lld", &n, &k);
      if (!k) {
        printf("1\n");
        continue;
      }
      long long ans = 0;
      if (n & 1) {
        ans = fastPower(fastPower(2, n - 1, 1000000007) + 1, k, 1000000007);
      } else {
        ans = fastPower(2, n - 1, 1000000007);
        for (long long i = 2; i <= k; ++i) {
          ans = (((fastPower(2, n - 1, 1000000007) - 1) * ans) % 1000000007 +
                 fastPower(2, (i - 1) * n, 1000000007)) %
                1000000007;
        }
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  