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 int mod = 1e9 + 7;
  int n, k;
  const int N = 200020;
  long long dp[N];
  int qpow(int x, int y) {
    int res = 1;
    while (y) {
      if (y & 1) res = 1ll * res * x % mod;
      x = 1ll * x * x % mod;
      y >>= 1;
    }
    return res;
  }
  void solve() {
    cin >> n >> k;
    long long ans = qpow(2, n - 1);
    if (n & 1) {
      int t1 = (qpow(2, n - 1) + 1) % mod;
      cout << qpow(t1, k) << endl;
    } else {
      long long res = 0;
      for (int i = 0; i < k; ++i) {
        res = (res +
               1ll * qpow(ans * 2 % mod, i) * qpow(ans - 1, k - i - 1) % mod) %
              mod;
      }
      res = (res + qpow(ans - 1, k)) % mod;
      cout << res << endl;
    }
  }
  int T = 1;
  signed main() {
    cin >> T;
    while (T--) solve();
    return 0;
  }
  