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 long long M = 1e9 + 7;
  const double pi = acos(-1.0);
  inline void read(long long &a) {
    a = 0;
    long long d = 1;
    char c;
    while (c = getchar(), c < '0' || c > '9')
      if (c == '-') d = -1;
    a = a * 10 + c - 48;
    while (c = getchar(), c >= '0' && c <= '9') a = a * 10 + c - 48;
    a *= d;
  }
  long long fpm(long long b, long long p) {
    long long ans = 1;
    while (p) {
      if (p & 1) ans = (ans * b) % M;
      p >>= 1;
      b = (b * b) % M;
    }
    return ans;
  }
  long long cc(long long n, long long m) {
    long long ans = 1;
    m = min(m, n - m);
    for (long long i = 0; i < m; i++) ans = (ans * (n - i) / (1 + i)) % M;
    return ans;
  }
  signed main() {
    long long t;
    read(t);
    while (t--) {
      long long n, k;
      read(n);
      read(k);
      if (k == 0) {
        printf("1\n");
        continue;
      }
      if (n % 2 == 0) {
        long long a = fpm(2, n);
        long long b = fpm(2, n - 1) - 1;
        long long ans = fpm(b, k);
        for (long long i = 1; i <= k; i++)
          ans = (ans + fpm(a, k - i) * fpm(b, i - 1)) % M;
        printf("%lld\n", ans);
      } else {
        long long b = 1 + fpm(2, n - 1);
        long long ans = fpm(b, k);
        printf("%lld\n", ans);
      }
    }
    return 0;
  }
  