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>
  const int mod = 1e9 + 7;
  using namespace std;
  long long jie[200005];
  long long qpow(long long a, long long b) {
    long long sum = 1, t = a;
    while (b) {
      if (b & 1) sum *= t;
      sum %= mod;
      t = t * t;
      t %= mod;
      b >>= 1;
    }
    return sum;
  }
  long long C(long long n, long long m) {
    return jie[n] * qpow(jie[m], mod - 2) % mod * qpow(jie[n - m], mod - 2) % mod;
  }
  int T;
  int main() {
    cin >> T;
    jie[0] = 1;
    for (int i = 1; i <= 200000; i++) {
      jie[i] = jie[i - 1] * i;
      jie[i] %= mod;
    }
    while (T--) {
      int n, k;
      cin >> n >> k;
      k;
      long long sum = 0, all = 0;
      long long ans = 0;
      for (int i = 0; i <= n; i += 2) sum += C(n, i), sum %= mod;
      for (int i = 0; i <= n; i++) all += C(n, i), all %= mod;
      if (k == 0) {
        cout << 1 << endl;
        continue;
      }
      if (n % 2) {
        cout << qpow(sum + 1, k) << endl;
      } else {
        for (int i = 1; i <= k + 1; i++) {
          ans += qpow(sum - 1, max(0, i - 1)) * qpow(all, max(k - i, 0));
          ans %= mod;
        }
        cout << ans << endl;
      }
    }
    return 0;
  }
  