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 N = 2e5 + 10, P = 1e9 + 7;
  int T, n, k;
  int dp[N];
  int power(int a, int b, int c = 1) {
    for (; b; b >>= 1, a = 1ll * a * a % P)
      if (b & 1) c = 1ll * c * a % P;
    return c;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    cin >> T;
    while (T--) {
      cin >> n >> k;
      memset(dp, 0, sizeof(dp));
      dp[0] = 1;
      int prod = 1, pw1 = power(2, n), pw2 = power(2, n - 1);
      for (int i = 1; i <= k; i++) {
        if (n % 2 == 0) {
          dp[i] = 1ll * dp[i - 1] * (pw2 - 1 + P) % P;
          dp[i] = (dp[i] + prod) % P;
        } else {
          dp[i] = 1ll * dp[i - 1] * (pw2 + 1) % P;
        }
        prod = 1ll * prod * pw1 % P;
      }
      cout << dp[k] << endl;
    }
    return 0;
  }
  