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;
  vector<long long> fact((long long)3e5);
  vector<long long> ifact((long long)3e5);
  long long modexp(long long x, long long y) {
    x %= (long long)1e9 + 7;
    if (x == 0) return 0;
    long long res = 1;
    while (y) {
      if (y & 1) {
        res = (x * res);
        res %= (long long)1e9 + 7;
      }
      y /= 2;
      x = (x * x);
      x %= (long long)1e9 + 7;
    }
    return res;
  }
  long long mod_inverse(long long n) { return modexp(n, (long long)1e9 + 7 - 2); }
  long long ncr(long long n, long long r) {
    long long ans = (fact[n] * ifact[r]);
    ans %= (long long)1e9 + 7;
    ans = (ans * ifact[n - r]);
    ans %= (long long)1e9 + 7;
    return ans;
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    if (k == 0) {
      cout << 1 << endl;
      return;
    }
    long long ans = 2;
    for (long long i = 2; i < n; i += 2) {
      ans += ncr(n, i);
    }
    if (n % 2) {
      ans = modexp(ans, k);
      ans %= (long long)1e9 + 7;
      cout << ans << endl;
      return;
    }
    long long total = modexp(ans - 1, k);
    for (long long i = 1; i <= k; i++) {
      long long curr = modexp(ans - 1, i - 1) * modexp(2, n * (k - i));
      curr %= (long long)1e9 + 7;
      total += curr;
      total %= (long long)1e9 + 7;
    }
    cout << total << endl;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    fact[0] = 1;
    for (long long i = 1; i < (long long)3e5; i++) {
      fact[i] = i * fact[i - 1];
      fact[i] %= (long long)1e9 + 7;
      ifact[i] = mod_inverse(fact[i]);
      ifact[i] %= (long long)1e9 + 7;
    }
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  