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;
  using li = long long;
  using ld = long double;
  using pii = pair<int, int>;
  const int INF = 1e9 + 13;
  const int N = 1e6 + 13;
  const int M = 1e9 + 7;
  const int B = 600;
  const int A = 256;
  const ld e = 1e-8;
  const int LOGN = 20;
  mt19937 rnd(time(0));
  int sum(int a, int b) {
    a += b;
    return (a >= M ? a - M : a);
  }
  int dif(int a, int b) { return sum(a, M - b); }
  int mul(int a, int b) { return a * 1ll * b % M; }
  int pow2(int n, int k) {
    if (k == 0) return 1;
    int pw = pow2(n, k / 2);
    pw = mul(pw, pw);
    return (k % 2 == 1 ? mul(pw, n) : pw);
  }
  int pw[N];
  void solve() {
    int n, k;
    cin >> n >> k;
    int ans = 0;
    int cur = 1;
    for (int i = k - 1; i >= 0; i--) {
      if (n % 2 == 0) {
        ans = sum(ans, mul(pow2(pw[i], n), cur));
        cur = mul(cur, dif(pw[n - 1], 1));
      } else {
        cur = mul(cur, sum(pw[n - 1], 1));
      }
    }
    cout << sum(ans, cur) << endl;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    pw[0] = 1;
    for (int i = 1; i < N; i++) pw[i] = mul(pw[i - 1], 2);
    int t;
    cin >> t;
    while (t--) solve();
  }
  