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;
  const int MAXN = 2e5 + 10, Mod = 1e9 + 7;
  int t, n, k;
  long long Pow(long long a, long long b) {
    a %= Mod;
    long long ans = 1;
    while (b) {
      if (b & 1) ans *= a, ans %= Mod;
      a *= a, a %= Mod;
      b >>= 1;
    }
    return ans;
  }
  long long jie[MAXN], ni[MAXN];
  void getall() {
    jie[0] = 1, ni[0] = 1;
    for (int i = 1; i < MAXN; ++i) jie[i] = (jie[i - 1] * i) % Mod;
    ni[MAXN - 1] = Pow(jie[MAXN - 1], Mod - 2);
    for (int i = MAXN - 2; i >= 1; --i) ni[i] = (ni[i + 1] * (i + 1)) % Mod;
  }
  long long calc(int a) {
    long long ans = jie[n];
    ans *= ni[a], ans %= Mod;
    ans *= ni[n - a], ans %= Mod;
    return ans;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin >> t;
    getall();
    while (t--) {
      cin >> n >> k;
      if (k == 0) {
        cout << 1 << endl;
        continue;
      }
      long long ans = 0, tmp1, tmp2, tt = Pow(2, k);
      if (n & 1) {
        tmp1 = 1;
        for (int i = 0; i <= n; i += 2) tmp1 += calc(i), tmp1 %= Mod;
        cout << Pow(tmp1, k) << endl;
      } else {
        tmp1 = 0, tmp2 = 1, ans = 0;
        for (int i = 0; i < n; i += 2) tmp1 += calc(i), tmp1 %= Mod;
        ans = Pow(tmp1, k);
        for (int i = 1; i <= k; ++i) {
          ans += tmp2 * Pow(2, 1ll * (k - i) * n);
          ans %= Mod;
          tmp2 *= tmp1;
          tmp2 %= Mod;
        }
        cout << ans << endl;
      }
    }
    return 0;
  }
  