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 <iostream>
  
  const int MOD = 1000000007;
  
  using namespace std;
  using ll = long long;
  
  ll pow(ll b, ll e) {
      if (e == 0) return 1;
      if (e == 1) return b;
      ll prev = pow(b, e / 2);
      return (((prev * prev) % MOD) * ((e % 2 == 1) ? (b) : (1))) % MOD;
  }
  
  ll solve(int n, int k, const ll& win, const ll& tie) {
      if (k == 0) return 1;
      // win
      ll w = (win * pow(pow(2, n), k - 1)) % MOD;
      // win later
      ll wl = (tie * solve(n, k - 1, win, tie)) % MOD;
      return (w + wl) % MOD;
  }
  
  void solve() {
      int n, k;
      cin >> n >> k;
      if (k == 0) {
          cout << 1 << endl;
          return;
      }
      ll kk = pow(2, n - 1);
      ll win, tie;
      if (n % 2 == 0) {
          win = 1;
          tie = kk - 1;
      } else if (n % 2 == 1) {
          win = 0;
          tie = kk + 1;
      }
      ll r = solve(n, k, win, tie);
      cout << r << endl;
  }
  
  int main() {
      int t;
      cin >> t;
      while (t--)
          solve();
  
      return 0;
  }