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;
  long long n, k;
  signed main() {
    long long T;
    scanf("%lld", &T);
    while (T--) {
      scanf("%lld%lld", &n, &k);
      long long w = 0, l = 0, b = 1;
      long long ta = 1, tb = 1, tc = 1;
      for (long long i = 0; i < n; i++) {
        tc <<= 1;
        tc %= 1000000007;
      }
      for (long long i = 1; i < n; i++) {
        ta <<= 1;
        ta %= 1000000007;
        tb <<= 1;
        tb %= 1000000007;
      }
      if (n & 1) {
        ta--;
        if (ta < 0) ta += 1000000007;
        tb--;
        if (tb < 0) tb += 1000000007;
      } else {
        tb -= 2;
        if (tb < 0) tb += 1000000007;
      }
      while (k--) {
        if (n & 1) {
          w *= tc;
          w %= 1000000007;
          b = (2 + tb) * b;
          b %= 1000000007;
        } else {
          w *= tc;
          w %= 1000000007;
          w += b;
          w %= 1000000007;
          b = (1 + tb) * b;
          b %= 1000000007;
        }
      }
      printf("%lld\n", (w + b) % 1000000007);
    }
    return 0;
  }
  