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 power(long long a, long long b, long long mod) {
    long long aInPow = a, ans = 1;
    while (b != 0) {
      if (b % 2) ans = (ans * aInPow) % mod;
      aInPow = (aInPow * aInPow) % mod;
      b >>= 1;
    }
    return ans;
  }
  void qs(vector<long long>& a, vector<long long>& b, long long l, long long r) {
    if (r - l <= 1) return;
    long long mid = a[l + rand() % (r - l - 1)], x = l, y = l;
    for (long long i = l; i < r; i++) {
      if (a[i] < mid) {
        swap(a[i], a[x]);
        swap(b[i], b[x]);
        if (x != y) {
          swap(a[i], a[y]);
          swap(b[i], b[y]);
        }
        x++;
        y++;
      } else if (a[i] == mid) {
        swap(a[i], a[y]);
        swap(b[i], b[y]);
        y++;
      }
    }
    qs(a, b, l, x);
    qs(a, b, y, r);
  }
  long long gcd(long long a, long long b) {
    while (b ^= a ^= b ^= a %= b)
      ;
    return a;
  }
  long long f(long long n) {
    long long i, ans = n;
    for (i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        while (n % i == 0) n /= i;
        ans -= ans / i;
      }
    }
    if (n > 1) ans -= ans / n;
    return ans;
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      long long p = 1e9 + 7;
      long long n, k;
      cin >> n >> k;
      long long c2 = power(2, n - 1, p) % p, c = (2 * c2) % p;
      if (n & 1)
        cout << power((c2 + 1) % p, k, p) << "\n";
      else
        cout << ((power(c, k, p) * 2 + c * power(c2 - 1, k, p)) % p *
                 power(c + 2, p - 2, p)) %
                    p
             << "\n";
    }
  }
  