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 long long N = 2e5 + 10, p = 1e9 + 7;
  inline long long ksm(long long a, long long b, const long long p) {
    long long c = 1;
    for (a %= p; b; a = a * a % p, b >>= 1)
      if (b & 1) c = c * a % p;
    return c;
  }
  void solve() {
    long long n, k, res = 0;
    cin >> n >> k;
    long long bs = ksm(2, n - 1, p), bs2 = bs * 2 % p;
    if (n & 1)
      cout << ksm(bs + 1, k, p) << '\n';
    else {
      for (long long x = k; x; --x) {
        long long a = ksm(bs + p - 1, k - x, p), b = ksm(bs2, x - 1, p);
        res = (res + a * b % p) % p;
      }
      res = (res + ksm(bs + p - 1, k, p)) % p;
      cout << res << '\n';
    }
  }
  signed main() {
    cout.tie(0)->sync_with_stdio(0);
    long long _;
    for (cin >> _; _--;) solve();
    return 0;
  }
  