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>
  using namespace std;
  const int mod = 1000000007;
  int i, j, k, n, m, x, y, t;
  int mi(int x, int y) {
    if (y == 0) return 1;
    int t = mi(x, y >> 1);
    ;
    t = (long long)t * t % mod;
    if (y & 1) t = (long long)t * x % mod;
    return t;
  }
  int main() {
    scanf("%d", &t);
    while (t--) {
      scanf("%d%d", &n, &m);
      int ans = 0;
      x = 1;
      for (i = 0; i < n; i += 2) {
        ans = (ans + x) % mod;
        x = (long long)x * (n - i) % mod * (n - i - 1) % mod *
            mi(i + 1, mod - 2) % mod * mi(i + 2, mod - 2) % mod;
      }
      if (n & 1) {
        ans++;
        printf("%d\n", mi(ans, m));
      } else {
        x = ans;
        ans = mi(ans, m);
        for (i = m; i >= 1; i--)
          ans =
              (ans + (long long)mi(mi(2, n), i - 1) * mi(x, (m - i)) % mod) % mod;
        printf("%d\n", ans);
      }
    }
    return 0;
  }
  