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;
  map<int, int> mp;
  long long f[1000010];
  const long long mod = 1000000007;
  void solve() {
    for (int i = 1; i <= 200010; i++) {
      f[i] = f[i - 1] * 2 % mod;
    }
  }
  long long sol(long long a, long long b) {
    long long ans = 1;
    while (b) {
      if (b & 1) {
        ans = a * ans % mod;
      }
      a = a * a % mod;
      b >>= 1;
    }
    return ans;
  }
  int main() {
    int t;
    cin >> t;
    f[0] = 1;
    solve();
    while (t--) {
      int n, k;
      cin >> n >> k;
      if (n & 1) {
        long long ans = sol(f[n - 1] + 1, k);
        cout << ans << endl;
        continue;
      }
      if (k == 0) {
        cout << 1 << endl;
        continue;
      }
      long long ans = sol(f[n - 1] - 1, k);
      for (int i = 1; i <= k; i++) {
        long long k1 = sol(f[n - 1] - 1, i - 1) * sol(f[n], k - i) % mod;
        ans = (ans + k1) % mod;
      }
      cout << ans << endl;
    }
    return 0;
  }
  