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;
  long long dp[200005];
  long long getmod(long long x) {
    long long tmp = 2;
    long long res = 1;
    while (x) {
      if (x & 1) res = (res * tmp) % 1000000007;
      x >>= 1;
      tmp = (tmp * tmp) % 1000000007;
    }
    return res;
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      dp[0] = 1;
      for (int i = 1; i <= k; i++) {
        if (n & 1) {
          dp[i] = (dp[i - 1] * getmod(n - 1) + dp[i - 1]) % 1000000007;
        } else {
          dp[i] =
              (dp[i - 1] * getmod(n - 1) - dp[i - 1] + 1000000007) % 1000000007 +
              getmod((long long)n * (i - 1));
          dp[i] %= 1000000007;
        }
      }
      cout << dp[k] << endl;
    }
    return 0;
  }
  