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;
  int const MAX = 2e5 + 5;
  int const MOD = 1e9 + 7;
  int powers[MAX];
  void precalc() {
    powers[0] = 1;
    for (int i = 1; i < MAX; ++i) {
      powers[i] = powers[i - 1] * 2;
      powers[i] %= MOD;
    }
  }
  void solve() {
    int n, k;
    cin >> n >> k;
    int sol = 1;
    if (k == 0) {
      cout << 1 << "\n";
      return;
    }
    if (n % 2 == 1) {
      for (int i = 0; i < k; ++i) {
        sol = ((long long)sol * (powers[n - 1] + 1)) % MOD;
      }
      cout << sol << "\n";
      return;
    }
    sol = 0;
    int pre[k];
    pre[0] = powers[n];
    for (int i = 1; i < k; ++i) {
      pre[i] = (((long long)powers[n] * pre[i - 1]) % MOD);
    }
    int aft[k];
    aft[0] = powers[n - 1] - 1;
    for (int i = 1; i < k; ++i) {
      aft[i] = (((long long)(powers[n - 1] - 1) * aft[i - 1]) % MOD);
    }
    sol += aft[k - 1];
    for (int i = 0; i < k; ++i) {
      long long total = 1;
      if (i > 0) {
        total *= pre[i - 1];
      }
      if (i < k - 1) {
        total *= aft[k - 2 - i];
      }
      total %= MOD;
      sol += total;
      sol %= MOD;
    }
    cout << sol << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    precalc();
    for (int i = 0; i < t; ++i) {
      solve();
    }
  }
  