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;
  void run() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  }
  const int N = 2e5 + 1;
  const int MOD = 1e9 + 7;
  long long fact[N], inv[N], invfact[N];
  void factInverse() {
    fact[0] = inv[1] = fact[1] = invfact[0] = invfact[1] = 1;
    for (long long i = 2; i < N; i++) {
      fact[i] = (fact[i - 1] * i) % MOD;
      inv[i] = MOD - (inv[MOD % i] * (MOD / i) % MOD);
      invfact[i] = (inv[i] * invfact[i - 1]) % MOD;
    }
  }
  int add(int a, int b) {
    if ((a += b) >= MOD)
      a -= MOD;
    else if (a < 0)
      a += MOD;
    return a;
  }
  long long mul(int x, int y) { return (1LL * x * y) % MOD; }
  long long nCr(int n, int r) {
    if (r > n) return 0;
    return mul(mul(fact[n], invfact[r]), invfact[n - r]);
  }
  int mem[N][2], vis[N][2], test_id;
  int cntEven, _2pwn, n;
  int dp(int i, bool equal) {
    if (i < 0) return 1;
    int& rt = mem[i][equal];
    if (vis[i][equal] == test_id) return rt;
    vis[i][equal] = test_id;
    if (equal == false) return rt = mul(_2pwn, dp(i - 1, 0));
    if (n & 1) return rt = mul(dp(i - 1, 1), add(cntEven, 1));
    return rt = add(mul(dp(i - 1, 1), cntEven), dp(i - 1, 0));
  }
  int main() {
    run();
    factInverse();
    int t;
    cin >> t;
    while (t--) {
      test_id++;
      int k;
      cin >> n >> k;
      cntEven = 0;
      for (int i = 0; i < n; i += 2) cntEven = add(cntEven, nCr(n, i));
      _2pwn = 1;
      for (int i = 0; i < n; i++) _2pwn = mul(_2pwn, 2);
      cout << dp(k - 1, 1) << "\n";
    }
  }
  