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 mod = 1000000007;
  int n, k;
  long long f[200005];
  long long ff;
  long long powlc(long long ac, long long bc) {
    if (bc == 0) return 1;
    if (bc == 1) return (ac % mod);
    long long cc = powlc(ac, bc / 2) % mod;
    cc = (cc * cc) % mod;
    if (bc % 2 == 1) cc = (cc * ac) % mod;
    return cc;
  }
  long long Cmod(int nx, int mx) {
    long long res =
        ((f[nx] * powlc(f[mx], mod - 2) % mod) * powlc(f[nx - mx], mod - 2)) %
        mod;
    return res;
  }
  void process() {
    f[0] = 1;
    for (int i = 1; i <= 200000; i++) {
      f[i] = (f[i - 1] * i) % mod;
    }
    int tcase = 1;
    cin >> tcase;
    while (tcase--) {
      cin >> n >> k;
      if (k == 0) {
        cout << 1 << endl;
        continue;
      }
      ff = 1;
      for (int i = 2; i <= n; i += 2) {
        ff = (ff + Cmod(n, i)) % mod;
      }
      if (n % 2 == 1)
        ff = (ff + 1) % mod;
      else
        ff = (ff - 1 + mod) % mod;
      long long res = 0;
      long long bounus = 1;
      for (int i = k; i >= 1; i--) {
        if (n % 2 == 0) {
          res = (res + (bounus * powlc(powlc(2, i - 1), n) % mod)) % mod;
        }
        bounus = (bounus * ff) % mod;
      }
      res = (res + bounus) % mod;
      cout << res << endl;
    }
  }
  int online = 2;
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    if (online == 0) {
      freopen("in.inp", "r", stdin);
      freopen("out.out", "w", stdout);
    } else if (online == 1) {
      freopen(".inp", "r", stdin);
      freopen(".out", "w", stdout);
    }
    process();
    return 0;
  }
  