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;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  void c_p_c() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  long long int n, k;
  long long int fact[200003];
  long long int inv[200003];
  long long int findInverse(long long int x) {
    long long int pow = 1000000007 - 2;
    long long int ans = 1;
    while (pow) {
      if (pow & 1) ans = (ans * x) % 1000000007;
      pow = pow >> 1;
      x = (x * x) % 1000000007;
    }
    return ans;
  }
  long long int pw(long long int x, long long int y) {
    long long int ans = 1;
    while (y) {
      if (y & 1) {
        ans = (ans * x) % 1000000007;
      }
      y = y >> 1;
      x = (x * x) % 1000000007;
    }
    return ans;
  }
  int32_t main() {
    c_p_c();
    fact[0] = 1;
    for (long long int i = 1; i < 200003; i++) {
      fact[i] = (fact[i - 1] * i) % 1000000007;
    }
    inv[200002] = findInverse(fact[200002]);
    for (long long int i = 200001; i > -0; i--) {
      inv[i] = (inv[i + 1] * (i + 1)) % 1000000007;
    }
    inv[0] = inv[1];
    long long int t;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      bool isEven = (n % 2 == 0) ? 1 : 0;
      long long int states = 0LL;
      long long int tot = 0;
      for (long long int i = 0; i < n; i++) {
        if (i % 2 == 0) {
          states = (states + (((fact[n] * inv[n - i]) % 1000000007) * (inv[i])) %
                                 1000000007) %
                   1000000007;
        }
        tot = (tot +
               (((fact[n] * inv[n - i]) % 1000000007) * (inv[i])) % 1000000007) %
              1000000007;
      }
      tot = (tot +
             (((fact[n] * inv[n - n]) % 1000000007) * (inv[n])) % 1000000007) %
            1000000007;
      long long int currState = 1LL;
      long long int ans = 0LL;
      for (long long int i = 0; i < k; i++) {
        if (isEven) {
          ans =
              (ans + (currState * pw(tot, k - i - 1)) % 1000000007) % 1000000007;
        }
        currState = (currState * ((!isEven + states) % 1000000007)) % 1000000007;
      }
      ans = (ans + currState) % 1000000007;
      cout << ans << "\n";
    }
    return 0;
  }
  