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>
  #pragma GCC optimize(3)
  using namespace std;
  inline int read() {
    int x = 0;
    int f = 0;
    char ch = 0;
    while (!isdigit(ch)) {
      f |= (ch == '-') ? 1 : 0;
      ch = getchar();
    }
    while (isdigit(ch)) {
      x = (x << 3) + (x << 1) + (ch ^ 48);
      ch = getchar();
    }
    return f ? -x : x;
  }
  const int mod = 1e9 + 7;
  int n, k, t;
  inline long long qpow(long long a, long long b) {
    long long ans = 1;
    while (b) {
      if (b & 1) ans = ans * a % mod;
      a = a * a % mod;
      b >>= 1;
    }
    return ans;
  }
  int main() {
    t = read();
    while (t--) {
      n = read();
      k = read();
      if (n & 1) {
        long long ans = qpow(2, n - 1) + 1;
        ans = qpow(ans, k);
        printf("%lld\n", ans);
      } else {
        long long temp = (qpow(2, n - 1) - 1 + mod) % mod;
        long long tmp = qpow(2, n);
        long long ans = qpow(temp, k);
        for (int i = 1; i <= k; ++i)
          ans = (ans + (long long)qpow(temp, k - i) * qpow(tmp, i - 1)) % mod;
        printf("%lld\n", ans);
      }
    }
  }
  