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 dp[1001];
  bool ready[1001];
  bool used[1001];
  long double dist(pair<long long, long long> p1, pair<long long, long long> p2) {
    return sqrt((p2.first - p1.first) * (p2.first - p1.first) +
                (p2.second - p1.second) * (p2.second - p1.second));
  }
  string repeat(long long t, string s) {
    string ans;
    for (long long i = 0; i < t; i++) {
      ans += s;
    }
    return ans;
  }
  long long bin_mod_exp(long long a, long long b) {
    if (b == 0) {
      return 1;
    }
    if (b == 1) {
      return a % 1000000007;
    }
    a %= 1000000007;
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = ((res % 1000000007) * a) % 1000000007;
      a = (a * a) % 1000000007;
      b >>= 1;
    }
    return res;
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    if (n == 1) {
      cout << bin_mod_exp(2LL, k) << endl;
      return;
    }
    if (!k) {
      cout << 1 << endl;
      return;
    }
    if (n % 2 == 0) {
      long long a = bin_mod_exp(bin_mod_exp(2LL, n - 1) + 1, 1000000007 - 2);
      long long ans = ((bin_mod_exp(2LL, n * k) -
                        bin_mod_exp(bin_mod_exp(2LL, n - 1) - 1, k)) *
                           a +
                       bin_mod_exp(bin_mod_exp(2LL, n - 1) - 1, k)) %
                      1000000007;
      while (ans < 0) {
        ans += 1000000007;
      }
      cout << ans << endl;
    } else {
      cout << bin_mod_exp(bin_mod_exp(2LL, n - 1) + 1, k) << endl;
    }
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  