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>
  #pragma GCC optimize("unroll-loops")
  #pragma GCC optimize("Ofast")
  #pragma GCC optimize("O0")
  #pragma GCC optimize("O1")
  using namespace std;
  const int mod = 1e9 + 7;
  const int N = 2e5 + 10;
  int n, k;
  vector<int> powers;
  void calc() {
    long long tmp = 1;
    for (int i = 0; i <= N; i++) {
      powers.push_back(tmp);
      tmp *= 2;
      tmp %= mod;
    }
  }
  void solve() {
    cin >> n >> k;
    long long tmp = powers[n - 1];
    long long res = 1;
    if (n % 2 == 1) {
      tmp++;
      for (int i = 0; i < k; i++) {
        res *= tmp;
        res %= mod;
      }
      cout << res << endl;
      return;
    }
    tmp--;
    for (int i = 0; i < k; i++) {
      res *= tmp;
      res %= mod;
    }
    long long tmp2[k + 10];
    tmp2[0] = 1;
    long long pn[k + 10];
    pn[0] = 1;
    for (int i = 1; i <= k; i++) {
      tmp2[i] = tmp2[i - 1] * tmp;
      tmp2[i] %= mod;
    }
    for (int i = 1; i <= k; i++) {
      pn[i] = pn[i - 1] * powers[n];
      pn[i] %= mod;
    }
    for (int i = 1; i <= k; i++) {
      res += tmp2[i - 1] * pn[k - i];
      res %= mod;
    }
    cout << res << endl;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    calc();
    while (t--) {
      solve();
    }
  }
  