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 power(long long a, long long b) {
    long long res = 1;
    a = a % 1000000007;
    while (b > 0) {
      if (b & 1) {
        res = (res * a) % 1000000007;
        b--;
      }
      a = (a * a) % 1000000007;
      b >>= 1;
    }
    return res;
  }
  long long fermat_inv(long long y) { return power(y, 1000000007 - 2ll); }
  long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t = 1;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      long long dp[k + 1];
      memset(dp, 0, sizeof(dp));
      dp[0] = 1ll;
      long long ans = 0ll;
      for (int i = 0; i < k; i++) {
        if (n % 2) {
          dp[i + 1] = (dp[i] * power(2ll, n - 1ll)) % 1000000007;
          dp[i + 1] = (dp[i + 1] + dp[i]) % 1000000007;
        } else {
          dp[i + 1] =
              (dp[i] * ((power(2ll, n - 1ll) - 1ll + 1000000007) % 1000000007)) %
              1000000007;
          ans = (ans + (dp[i] * power(power(2ll, k - i - 1), n)) % 1000000007) %
                1000000007;
        }
      }
      ans = (ans + dp[k]) % 1000000007;
      cout << ans << "\n";
    }
    return 0;
  }
  