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>
  const int MOD = 1e9 + 7;
  long long BinPow(int x, int power) {
    if (power == 0) {
      return 1;
    } else if (power == 1) {
      return x % MOD;
    } else {
      long long s = BinPow(x, power / 2);
      if (power % 2) {
        return (s * s % MOD) * x % MOD;
      }
      return s * s % MOD;
    }
  }
  void Solve() {
    int n = 0, k = 0;
    std::cin >> n >> k;
    std::vector<long long> pow(n + k + 1);
    pow[0] = 1;
    for (int i = 1; i < pow.size(); ++i) {
      pow[i] = pow[i - 1] * 2 % MOD;
    }
    long long ans = 0;
    if (n % 2 == 0) {
      for (int i = 0; i < k; ++i) {
        ans += BinPow(pow[n - 1] - 1, k - 1 - i) * BinPow(pow[n], i) % MOD;
        ans %= MOD;
      }
      ans += BinPow(pow[n - 1] - 1, k);
      ans %= MOD;
    } else {
      ans = BinPow(pow[n - 1] + 1, k);
    }
    std::cout << ans << '\n';
  }
  int main() {
    int t = 0;
    std::cin >> t;
    while (t--) {
      Solve();
    }
  }
  