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;
  int n, k, T, mod = 1e9 + 7;
  long long ans = 0, inv[200010], frac[200010], res;
  long long qpow(long long x, long long y) {
    long long ans = 1;
    while (y) {
      if (y & 1) ans = ans * x % mod;
      x = x * x % mod;
      y >>= 1;
    }
    return ans;
  }
  long long C(int x, int y) { return frac[x] * inv[y] % mod * inv[x - y] % mod; }
  int main() {
    cin >> T;
    frac[0] = inv[0] = 1;
    for (int i = 1; i <= 200000; i++) frac[i] = frac[i - 1] * i % mod;
    for (int i = 1; i <= 200000; i++) inv[i] = qpow(frac[i], mod - 2);
    while (T--) {
      cin >> n >> k;
      ans = res = 0;
      for (int i = 0; i < n; i += 2) ans = (ans + C(n, i)) % mod;
      if (n & 1) {
        ans++;
        printf("%lld\n", qpow(ans, k) % mod);
      } else {
        res = qpow(ans, k);
        for (int i = 1; i <= k; i++)
          res = (res + qpow(qpow(2, n), k - i) * qpow(ans, i - 1)) % mod;
        printf("%lld\n", res);
      }
    }
  }
  