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 int pow_cal[200001];
  long long int power(long long int a, long long int b) {
    long long int res = 1;
    while (b > 0) {
      if (b & 1) res = (res * a) % 1000000007;
      a = (a * a) % 1000000007;
      b /= 2;
    }
    return res % 1000000007;
  }
  long long int abc(long long int n, long long int k) {
    if (n == 1) return pow_cal[k];
    if (k == 0) return 1;
    if (n & 1) {
      long long int x = (pow_cal[n - 1] + 1) % 1000000007;
      x = power(x, k);
      return x;
    } else {
      if (k == 1) return pow_cal[n - 1];
      long long int x = power(pow_cal[n], k - 1);
      x = x + ((pow_cal[n - 1] - 1) * (abc(n, k - 1))) % 1000000007;
      x %= 1000000007;
      return x;
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    pow_cal[0] = 1;
    for (int i = 1; i < 200001; i++) {
      pow_cal[i] = (2 * pow_cal[i - 1]) % 1000000007;
    }
    int T;
    cin >> T;
    while (T--) {
      long long int n, k;
      cin >> n >> k;
      cout << abc(n, k) << '\n';
    }
  }
  