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;
  long long mod = 1e9 + 7;
  const int mx = 2e5 + 5;
  vector<int> fact(mx, 1), inv(mx, 1);
  long long binpow(long long a, long long b, long long m) {
    a %= m;
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % m;
      a = a * a % m;
      b >>= 1;
    }
    return res;
  }
  void precompute() {
    fact[0] = inv[0] = 1;
    for (int i = 1; i < mx; i++) {
      fact[i] *= fact[i - 1] % mod;
      inv[i] = binpow(i, mod - 2, mod);
    }
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    int flag = 0;
    if (n % 2 == 1) flag = 1;
    long long eq = 1, ans = 0;
    for (int i = 1; i <= k; i++) {
      if (flag == 1) {
        eq *= binpow(2, n - 1, mod) + 1;
        eq %= mod;
      } else {
        long long gr;
        if (i == k)
          gr = 1;
        else
          gr = binpow(binpow(2, n, mod) % mod, k - i, mod);
        gr %= mod;
        gr *= eq;
        gr %= mod;
        ans += gr;
        eq *= binpow(2, n - 1, mod) - 1;
        eq += mod;
        eq %= mod;
      }
    }
    ans += eq;
    ans %= mod;
    cout << ans << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout << fixed << setprecision(10);
    int t = 1;
    cin >> t;
    precompute();
    while (t--) solve();
  }
  