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 mod = 1e9 + 7;
  int mod_pow(int b, int p) {
    int ans = 1;
    while (p) {
      if (p & 1) {
        ans = (1ll * ans * b) % mod;
        p--;
      }
      b = (1ll * b * b) % mod;
      p /= 2;
    }
    return ans;
  }
  const int N = (2e5 + 5);
  int fac[N], inv_fac[N];
  void calc_fac() {
    fac[0] = 1;
    inv_fac[0] = 1;
    for (int i = 1; i < N; i++) {
      fac[i] = (1ll * fac[i - 1] * i) % mod;
      inv_fac[i] = mod_pow(fac[i], mod - 2);
    }
  }
  void solve() {
    int n, k;
    cin >> n >> k;
    if (k == 0) {
      cout << 1;
      cout << '\n';
      return;
    }
    int e = 0;
    for (int i = 0; i < n; i += 2) {
      e = (e + 1ll * fac[n] * inv_fac[n - i] % mod * inv_fac[i] % mod) % mod;
    }
    if (n & 1) e = (e + 1) % mod;
    if (n & 1) {
      int ans = mod_pow(e, k);
      cout << ans;
      cout << '\n';
      return;
    }
    int p = mod_pow(2, n);
    int pp = mod_pow(p, k);
    int ans = 0;
    int left = 1;
    int right = pp;
    for (int i = 0; i < k; i++) {
      right = 1ll * right * mod_pow(p, mod - 2) % mod;
      ans = (ans + 1ll * left * right) % mod;
      left = (1ll * left * e) % mod;
    }
    ans = (ans + 1ll * left) % mod;
    cout << ans;
    cout << '\n';
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    calc_fac();
    int tc;
    cin >> tc;
    while (tc--) solve();
    return 0;
  }
  