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 _MAX = 1e+9 + 7;
  int _pow(int n, int __pow, vector<int>* _mem = nullptr) {
    int res = 1;
    if (_mem != nullptr) _mem->emplace_back(res);
    while (__pow--) {
      res = ((long long)res * n) % _MAX;
      if (_mem != nullptr) _mem->emplace_back(res);
    }
    return res;
  }
  int solve(int n, int k) {
    if (!k) return 1;
    int _f = _pow(2, n - 1), _g = _pow(2, n) - 1;
    if (n % 2) return _pow(((long long)_f + 1) % _MAX, k);
    _f--;
    vector<int> _f_mem, _g_mem;
    int res = _pow(_f, k, &_f_mem);
    _pow((1 + (long long)_g) % _MAX, k - 1, &_g_mem);
    for (int i = 0; i <= k - 1; i++)
      res = ((((long long)_f_mem[i] * _g_mem[k - 1 - i]) % _MAX) + res) % _MAX;
    return res;
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      cout << solve(n, k) << endl;
    }
  }
  