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>
  long long int power(long long int a, long long int b) {
    long long int res = 1;
    a %= 1000000007;
    assert(b >= 0);
    for (; b; b >>= 1) {
      if (b & 1) res = res * a % 1000000007;
      a = a * a % 1000000007;
    }
    return res;
  }
  using namespace std;
  const long double PI = acos(-1);
  const long double EPS = 1e-9;
  const long long int INF = 1e18;
  long long int gcd(long long int a, long long int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  void solve() {
    long long int n, k, i;
    cin >> n >> k;
    long long int ans00[k + 1];
    long long int ans01[k + 1];
    long long int ans10[k + 1];
    long long int ans11[k + 1];
    long long int ans[k + 1];
    long long int answ[k + 1];
    answ[0] = 1;
    ans[0] = 1;
    long long int fo = (power(2, n - 1) - 1) % 1000000007;
    long long int fe = (power(2, n - 1)) % 1000000007;
    for (i = 1; i <= k; i++) {
      answ[i] = (answ[i - 1] * (2 * fe)) % 1000000007;
      if (n % 2) {
        ans11[i] = ans[i - 1];
      } else
        ans11[i] = 0;
      if ((n % 2) == 0) {
        ans10[i] = answ[i - 1];
      } else
        ans10[i] = 0;
      if (n % 2) {
        ans00[i] = (((ans[i - 1]) * (fo + 1)) % 1000000007);
      } else {
        ans00[i] = (((ans[i - 1]) * (fe - 1)) % 1000000007);
      }
      ans[i] = ans00[i] + ans10[i] + ans11[i];
      ans[i] = (ans[i] % 1000000007);
    }
    cout << ans[k] << endl;
    return;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int TC = 1, te = 0;
    cin >> TC;
    while (te++ < TC) {
      solve();
    }
    cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
         << "ms\n";
    return 0;
  }
  