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;
  inline int inc(int x, int y) { return x += y - mod, x += x >> 31 & mod; }
  inline int dec(int x, int y) { return x -= y, x += x >> 31 & mod; }
  inline int mul(int x, int y) { return long long(x) * y % mod; }
  inline int sed(int x, int y) { return y ? x & 1 ? mod - y : y : 0; }
  int qpow(int a, int b) {
    b += b >> 31 & (mod - 1);
    int c = 1;
    for (; b; b >>= 1, a = mul(a, a)) {
      if (b & 1) c = mul(a, c);
    }
    return c;
  }
  const int maxn = 2e5 + 5;
  int T, n, k;
  int main() {
    scanf("%d", &T);
    while (T--) {
      scanf("%d %d", &n, &k);
      int a = qpow(2, n);
      int A = qpow(a, k);
      int I = qpow(a, mod - 2);
      int c = 1;
      int s = 0;
      for (int i = int(0); i <= int(k - 1); i++) {
        int p = dec((mod + 1) / 2, (n & 1 ? I : 0));
        int q = n & 1 ? 0 : I;
        s = inc(s, mul(c, p));
        c = mul(c, dec(1, inc(p, q)));
      }
      printf("%d\n", mul(dec(1, s), A));
    }
    return 0;
  }
  