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 int gcd(long long int a, long long int b) {
    if (a == 0) return b;
    if (b == 0) return a;
    return gcd(b % a, a);
  }
  long long int my_pow(long long int a, long long int b) {
    if (b == 0) return 1;
    if (b == 1) return a;
    long long int half = my_pow(a, b / 2);
    long long int ans = (half * half);
    if (b % 2) {
      ans = (ans * a);
    }
    return ans;
  }
  long long int my_pow_mod_p(long long int a, long long int b, long long int p) {
    a = (a % p);
    if (b == 0) return 1;
    if (b == 1) return a % p;
    long long int half = my_pow_mod_p(a, b / 2, p);
    long long int ans = (half * half) % p;
    if (b % 2) {
      ans = (ans * a) % p;
    }
    return ans;
  }
  class combinatorics {
    long long int N;
    long long int P;
    vector<long long int> fact;
    void cal() {
      fact[0] = 1;
      fact[1] = 1;
      for (long long int i = 2; i < N; i++) {
        fact[i] = (i * fact[i - 1]) % P;
      }
    }
  
   public:
    combinatorics(long long int n1, long long int p1) {
      P = p1;
      N = max(n1 + 1, 2LL);
      fact.resize(N);
      cal();
    }
    long long int nPr(long long int n, long long int r) {
      if (n < r) {
        return 0;
      }
      long long int val = (fact[n] * my_pow_mod_p(fact[n - r], P - 2, P)) % P;
      return val;
    }
    long long int nCr(long long int n, long long int r) {
      if (n < r) {
        return 0;
      }
      long long int val = nPr(n, r);
      val = (val * my_pow_mod_p(fact[r], P - 2, P)) % P;
      return val;
    }
    long long int get_fact(long long int n) { return fact[n]; }
  };
  void print_vec(vector<long long int> &V, string name) {
    cout << name << " : " << endl;
    long long int n = V.size();
    for (long long int i = 0; i < n; i++) {
      cout << V[i] << " ";
    }
    cout << endl;
  }
  void handle_even(long long int n, long long int k) {
    long long int eq_val = my_pow_mod_p(2LL, n - 1, 1000000007);
    eq_val = (eq_val - 1 + 1000000007) % 1000000007;
    long long int gr_val = 1LL;
    long long int a = eq_val;
    long long int b = gr_val;
    long long int ans = (a + b) % 1000000007;
    for (long long int i = 1; i < k; i++) {
      long long int val1 = my_pow_mod_p(2LL, i, 1000000007);
      long long int val2 = my_pow_mod_p(val1, n, 1000000007);
      long long int curr_ans = val2;
      long long int val3 = (eq_val * ans) % 1000000007;
      ans = (curr_ans + val3) % 1000000007;
    }
    cout << ans << endl;
  }
  void handle_odd(long long int n, long long int k) {
    long long int val = my_pow_mod_p(2LL, n - 1, 1000000007);
    val = (val + 1) % 1000000007;
    val = my_pow_mod_p(val, k, 1000000007);
    cout << val << endl;
  }
  void solve() {
    long long int n, k;
    cin >> n >> k;
    if (k == 0) {
      cout << 1 << endl;
      return;
    }
    if (n % 2 == 0)
      handle_even(n, k);
    else
      handle_odd(n, k);
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    long long int T = 1;
    cin >> T;
    while (T--) {
      solve();
    }
    return 0;
  }
  