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;
  using ll = long long;
  const int mod = 1e9 + 7;
  const int nax = 2e5;
  int T, n, k;
  ll qexp(int a, int b) {
    ll res = 1;
    while (b) {
      if (b & 1) res = res * a % mod;
      a = (ll)a * a % mod;
      b >>= 1;
    }
    return res;
  }
  void solve() {
    cin >> n >> k;
    ll ans = 1, pre = 1;
    if (n & 1) {
      for (int i = k - 1; ~i; i -= 1) ans = ans * (1 + qexp(2, n - 1)) % mod;
    } else {
      ans = 0;
      for (int i = k - 1; ~i; i -= 1) {
        ans = (ans + pre * qexp(qexp(2, n), i) % mod) % mod;
        pre = pre * (qexp(2, n - 1) - 1 + mod) % mod;
      }
      ans = (ans + pre) % mod;
    }
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> T;
    while (T--) solve();
    return 0;
  }
  