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 binpow(long long a, long long b, long long m) {
    a %= m;
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % m;
      a = a * a % m;
      b >>= 1;
    }
    return res;
  }
  int main() {
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      long long tot = binpow(2, k, (long long)1000000007);
      tot = binpow(tot, n, (long long)1000000007);
      for (int i = 0; i < k; i++) {
        if (n % 2) {
          long long g = binpow(2, n - 1, (long long)1000000007);
          tot = (tot -
                 (((g - 1) * binpow(g + 1, i, (long long)1000000007)) %
                  (long long)1000000007 *
                  binpow(2 * g, k - i - 1, (long long)1000000007)) %
                     (long long)1000000007 +
                 (long long)1000000007) %
                (long long)1000000007;
        } else {
          long long g = binpow(2, n - 1, (long long)1000000007);
          tot = (tot -
                 ((g * binpow(g - 1, i, (long long)1000000007)) %
                  (long long)1000000007 *
                  binpow(2 * g, k - i - 1, (long long)1000000007)) %
                     (long long)1000000007 +
                 (long long)1000000007) %
                (long long)1000000007;
        }
      }
      cout << tot << "\n";
    }
    return 0;
  }
  