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 luth[200005];
  class game {
   private:
    int n, k;
  
   public:
    long long dp[200005][5];
    long long res = 0;
    void in() {
      int x, y;
      std::cin >> x >> y;
      n = x;
      k = y;
    }
    void solve() {
      dp[0][0] = luth[n];
      for (int i = 1; i <= k; i++) {
        dp[i][0] = (dp[i - 1][0] * luth[n]) % 1000000007;
      }
      if (k == 0) {
        res = 1;
        return;
      }
      if (k == 1) {
        if (n % 2 == 1) {
          res = ((luth[n - 1]) + 1) % 1000000007;
        } else {
          res = (luth[n - 1]) % 1000000007;
        }
        return;
      }
      if (n % 2 == 1) {
        dp[0][1] = (luth[n - 1] + 1) % 1000000007;
        for (int i = 1; i < k; i++) {
          dp[i][1] = (dp[i - 1][1] * (luth[n - 1] + 1)) % 1000000007;
        }
      } else {
        dp[0][1] = (luth[n - 1]) % 1000000007;
        for (int i = 1; i < k; i++) {
          dp[i][1] =
              ((dp[i - 1][1] * (luth[n - 1] - 1)) + dp[i - 1][0]) % 1000000007;
        }
      }
      res = dp[k - 1][1];
    }
    void out() { std::cout << res << std::endl; }
  };
  int main() {
    int tests;
    std::cin >> tests;
    luth[0] = 1;
    for (int i = 1; i <= 200003; i++) {
      luth[i] = luth[i - 1] * 2;
      if (luth[i] >= 1000000007) luth[i] = luth[i] % 1000000007;
    }
    for (int i = 0; i < tests; i++) {
      game *obj = new game;
      obj->in();
      obj->solve();
      obj->out();
    }
    return 0;
  }
  