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;
  long long read() {
    long long x = 0, y = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '-') y = -1;
      ch = getchar();
    }
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return x * y;
  }
  long long b[100005];
  long long c[100005];
  const int mod = 1e9 + 7;
  long long qpow(long long x, long long y) {
    long long res = 1, m = x;
    while (y) {
      if (y & 1) res = (res * m) % mod;
      m = (m * m) % mod;
      y >>= 1;
    }
    return res;
  }
  int main() {
    int T = read();
    while (T--) {
      long long n = read(), k = read();
      if (n & 1) {
        long long ans = qpow(2, n - 1);
        ans = (ans + 1) % mod;
        ans = qpow(ans, k);
        printf("%lld\n", ans);
      } else {
        long long ans = 1;
        long long m = 1;
        for (int i = 1; i <= k; i++) {
          ans = ((ans * (qpow(2, n - 1) - 1)) % mod + m) % mod;
          m = (m * qpow(2, n)) % mod;
        }
        ans = ((ans % mod) + mod) % mod;
        printf("%lld\n", ans);
      }
    }
    return 0;
  }
  