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 long long int MOD = 1e9 + 7;
  long long int binpow(long long int a, long long int b) {
    long long int r = 1;
    while (b) {
      if (b & 1) r = (r % MOD * a % MOD) % MOD;
      a = (a % MOD * a % MOD) % MOD;
      b /= 2;
    }
    return r;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int t = 1, testcases = 1;
    cin >> t;
    while (t >= testcases) {
      long long int n, k;
      cin >> n >> k;
      long long int ans = 0;
      if (n % 2) {
        ans = (binpow(2, n - 1) + 1) % MOD;
        ans = binpow(ans, k);
      } else {
        long long int dp[k + 1];
        dp[0] = 1;
        long long int c = (binpow(2, n - 1) - 1 + MOD) % MOD;
        for (long long int i = 1; i <= k; i++) {
          long long int p = n * (i - 1);
          long long int o = binpow(2, p);
          dp[i] = (o % MOD + (c % MOD * dp[i - 1] % MOD) % MOD) % MOD;
        }
        ans = dp[k];
      }
      cout << ans << "\n";
      testcases++;
    }
    cerr << "Time Taken:" << (float)clock() / CLOCKS_PER_SEC * 1000 << "ms"
         << "\n";
    return 0;
  }
  