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 INF = 0x3f3f3f3f;
  const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
  const long long MOD = 1e9 + 7;
  const int MAX = 2e5 + 10;
  long long aux[MAX][2];
  long long dp[MAX][2];
  int N, K;
  long long fexp(long long a, long long e) {
    long long ans = 1;
    while (e) {
      if (e & 1) ans = ans * a % MOD;
      a = a * a % MOD;
      e >>= 1;
    }
    return ans;
  }
  long long solve(int pos, bool won) {
    if (pos == K) return 1;
    long long &ans = dp[pos][won];
    if (ans != -1) return ans;
    ans = 0;
    if (won) {
      ans = (ans + fexp(2, N) * solve(pos + 1, 1) % MOD) % MOD;
    } else {
      ans = (ans + solve(pos + 1, !(N & 1))) % MOD;
      ans = (ans + (aux[N][0] + MOD - !(N & 1)) * solve(pos + 1, 0) % MOD) % MOD;
    }
    return ans;
  }
  inline void test_case() {
    cin >> N >> K;
    memset(dp, -1, sizeof dp);
    cout << solve(0, 0) << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    aux[0][0] = 1;
    for (int i = 1; i < MAX; i++) {
      for (int k : {0, 1}) {
        aux[i][k] = (aux[i - 1][k] + aux[i - 1][not k]) % MOD;
      }
    }
    int number_of_tests;
    cin >> number_of_tests;
    while (number_of_tests--) {
      test_case();
    }
    return 0;
  }
  