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;
  inline void rd(char &x) { scanf("%c", &x); }
  inline void rd(char *s) { scanf("%s", s); }
  inline void rd(int &x) { scanf("%d", &x); }
  inline void rd(long long &x) { scanf("%lld", &x); }
  template <typename T, typename... U>
  inline void rd(T &a, U &...b) {
    rd(a), rd(b...);
  }
  template <typename T>
  inline void write(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
  }
  const int maxn = 214514;
  const long long mod = 1e9 + 7;
  long long pw[maxn] = {1};
  long long n, k;
  long long qpow(long long a, long long b) {
    b %= (mod - 1);
    long long ret = 1;
    while (b) {
      if (b & 1) ret = ret * a % mod;
      a = a * a % mod, b >>= 1;
    }
    return ret;
  }
  long long solve(int k) {
    long long ret = 1;
    for (int i = 0, __i = k - 1; i <= __i; ++i) {
      if (n & 1) {
        ret = (1 + pw[n - 1]) * ret % mod;
      } else {
        ret = qpow(2, i * n) + (pw[n - 1] - 1) * ret % mod;
        ret = (ret + mod) % mod;
      }
    }
    return ret;
  }
  int main() {
    for (int i = 1, __i = 200100; i <= __i; ++i) pw[i] = pw[i - 1] * 2 % mod;
    int T;
    rd(T);
    for (int kase = 1, __kase = T; kase <= __kase; ++kase) {
      rd(n, k);
      printf("%lld\n", solve(k));
    }
  }
  