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;
  template <typename T>
  inline void read(T &n) {
    T w = 1;
    n = 0;
    char ch = getchar();
    while (!isdigit(ch) && ch != EOF) {
      if (ch == '-') w = -1;
      ch = getchar();
    }
    while (isdigit(ch) && ch != EOF) {
      n = (n << 1) + (n << 3) + (ch & 15);
      ch = getchar();
    }
    n *= w;
  }
  const int N = 2e5 + 5;
  int T, n, k;
  long long a[N + 2], ia[N + 2], b[N + 2], s[N + 2];
  long long ksm(long long a, long long b) {
    long long x = 1;
    while (b) {
      if (b & 1) x = x * a % 1000000007;
      a = a * a % 1000000007;
      b >>= 1;
    }
    return x;
  }
  long long dp(int i) {
    if (i > k) return 1;
    long long res = (s[(n - 1) >> 1] + (n & 1)) * dp(i + 1) % 1000000007;
    return (res + (n % 2 == 0) * ksm(b[k - i], n)) % 1000000007;
  }
  int main() {
    read(T);
    a[0] = ia[0] = 1;
    for (int i = 1; i < N; i++) {
      a[i] = a[i - 1] * i % 1000000007;
      ia[i] = ksm(a[i], 1000000007 - 2);
    }
    b[0] = 1;
    for (int i = 1; i < N; i++) b[i] = b[i - 1] * 2 % 1000000007;
    while (T--) {
      read(n);
      read(k);
      s[0] = 1;
      for (int i = 1; (i << 1) <= n; i++)
        s[i] = (s[i - 1] +
                a[n] * ia[n - (i << 1)] % 1000000007 * ia[i << 1] % 1000000007) %
               1000000007;
      printf("%lld\n", dp(1));
    }
    return 0;
  }
  