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;
  long long mod = 1e9 + 7;
  long long dp1[200005][2];
  long long dp3[200005];
  long long f1(long long i, long long j) {
    if (dp1[i][j] != -1) return dp1[i][j];
    if (i == 0) {
      if (j == 0)
        return dp1[i][j] = 1;
      else
        return dp1[i][j] = 0;
    }
    return dp1[i][j] = (f1(i - 1, j) + f1(i - 1, (j + 1) % 2)) % mod;
  }
  long long f2(vector<long long> &dp2, long long i, long long n) {
    if (i == 0) {
      return dp3[i] = 1;
    }
    if (dp3[i] != -1) return dp3[i];
    long long ans = 0;
    if (n % 2) {
      ans += f2(dp2, i - 1, n);
      ans %= mod;
    } else {
      ans += dp2[i - 1];
      ans %= mod;
    }
    long long d = dp1[n][0];
    if (n % 2 == 0) {
      d = (d - 1 + mod) % mod;
    }
    ans += d * f2(dp2, i - 1, n) % mod;
    ans %= mod;
    return dp3[i] = ans;
  }
  void solvef() {
    long long n, k;
    cin >> n >> k;
    for (int i = 0; i < 200005; i++) dp1[i][0] = dp1[i][1] = -1;
    f1(n, 0);
    if (k == 0) {
      cout << 1 << endl;
      return;
    }
    long long pn = 1;
    for (int i = 0; i < n; i++) pn = pn * 2 % mod;
    vector<long long> dp2(k + 1);
    dp2[0] = 1;
    for (int i = 1; i <= k; i++) {
      dp2[i] = dp2[i - 1] * pn % mod;
    }
    for (int i = 0; i < 200005; i++) dp3[i] = -1;
    long long ans = f2(dp2, k, n);
    cout << ans << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t;
    cin >> t;
    while (t--) solvef();
    return 0;
  }
  