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 extEuclid(long long a, long long b, long long &x, long long &y) {
    long long xx = y = 0;
    long long yy = x = 1;
    while (b) {
      long long q = a / b;
      tie(a, b) = make_tuple(b, a % b);
      tie(x, xx) = make_tuple(xx, x - q * xx);
      tie(y, yy) = make_tuple(yy, y - q * yy);
    }
    return a;
  }
  long long modInverse(long long b, long long m) {
    long long x, y;
    long long d = extEuclid(b, m, x, y);
    if (d != 1) return -1;
    return (x + m) % m;
  }
  const int maxN = 200010;
  long long MOD = 1e9 + 7, fact[maxN], invFact[maxN];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    fact[0] = 1;
    invFact[0] = 1;
    for (int i = 1; i < maxN; i++) {
      fact[i] = (i * fact[i - 1]) % MOD;
      invFact[i] = modInverse(fact[i], MOD);
    }
    int tt;
    cin >> tt;
    while (tt--) {
      int n, k;
      cin >> n >> k;
      long long combis = 0;
      for (int i = 0; i < n; i += 2) {
        long long ncr = (fact[n] * invFact[i]) % MOD;
        ncr = (ncr * invFact[n - i]) % MOD;
        combis = (combis + ncr) % MOD;
      }
      long long _2pwn = 1;
      for (int i = 0; i < n; i++) _2pwn = (_2pwn * 2) % MOD;
      long long ans = 1;
      if (n & 1) {
        combis++;
        for (int i = 0; i < k; i++) ans = (ans * combis) % MOD;
      } else {
        for (int i = 0; i < k; i++) ans = (ans * combis) % MOD;
        long long pow2pwn[k];
        pow2pwn[0] = 1;
        for (int i = 1; i < k; i++) pow2pwn[i] = (pow2pwn[i - 1] * _2pwn) % MOD;
        long long front = 1;
        for (int i = 0; i < k; i++) {
          ans += (front * pow2pwn[k - i - 1]) % MOD;
          ans %= MOD;
          front = (front * combis) % MOD;
        }
      }
      cout << ans << endl;
    }
    return 0;
  }
  