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 MOD = 1e9 + 7;
  const int N = 3e5 + 5;
  int f[N], inv[N], n, k, XOR, dp[N][2];
  int pow(int a, int b) {
    int res = 1;
    while (b) {
      if (b % 2) {
        res = 1ll * res * a % MOD;
      }
      a = 1ll * a * a % MOD;
      b /= 2;
    }
    return res;
  }
  void precompute() {
    f[1] = 1;
    for (int i = 2; i < N; ++i) {
      f[i] = 1ll * f[i - 1] * i % MOD;
    }
    inv[N - 1] = pow(f[N - 1], MOD - 2);
    for (int i = N - 2; i >= 0; --i) {
      inv[i] = 1ll * inv[i + 1] * (i + 1) % MOD;
    }
  }
  int nCr(int x, int y) {
    return x - y > 0 ? 1ll * f[x] * inv[y] % MOD * inv[x - y] % MOD : 0;
  }
  int go(int bit, bool equal) {
    if (bit == k) {
      return 1;
    }
    if (dp[bit][equal] != -1) {
      return dp[bit][equal];
    }
    int cnt = 0;
    if (equal == 0) {
      cnt = (cnt + 1ll * pow(2, n) * go(bit + 1, equal)) % MOD;
    } else {
      if (n % 2 == 0) {
        cnt = (cnt + go(bit + 1, false));
      }
      if (n % 2) {
        cnt = (cnt + go(bit + 1, true)) % MOD;
      }
      cnt = (cnt + 1ll * XOR * go(bit + 1, true)) % MOD;
    }
    return dp[bit][equal] = cnt;
  }
  void solve() {
    cin >> n >> k;
    memset(dp, -1, sizeof(dp));
    XOR = 0;
    for (int i = 0; i < n; i += 2) {
      XOR = (XOR + nCr(n, i)) % MOD;
    }
    cout << go(0, true) << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    precompute();
    int t = 1;
    cin >> t;
    int i = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  