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>
  namespace FastRead {
  char __buff[5000];
  long long __lg = 0, __p = 0;
  char nc() {
    if (__lg == __p) {
      __lg = fread(__buff, 1, 5000, stdin);
      __p = 0;
      if (!__lg) return EOF;
    }
    return __buff[__p++];
  }
  template <class T>
  void read(T& __x) {
    T __sgn = 1;
    char __c;
    while (!isdigit(__c = nc()))
      if (__c == '-') __sgn = -1;
    __x = __c - '0';
    while (isdigit(__c = nc())) __x = __x * 10 + __c - '0';
    __x *= __sgn;
  }
  }  // namespace FastRead
  using namespace FastRead;
  using namespace std;
  const long long N = 1e5 + 10;
  const long long M = 1e9 + 7;
  const long double PI = acos(-1);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  long long n, k;
  long long pw(long long n, long long p) {
    long long ans = 1;
    for (; p; p >>= 1, n = n * n % M)
      if (p & 1) ans = ans * n % M;
    return ans;
  }
  void Test() {
    cin >> n >> k;
    if (n & 1) return cout << pw((pw(2, n - 1) + 1), k) << '\n', void();
    long long X = (pw(2, n - 1) - 1 + M) % M;
    long long ans = pw(X, k);
    for (long long c = 1; c <= k; c++)
      ans = (ans + pw(X, c - 1) * pw(pw(2, n), k - c)) % M;
    cout << ans << '\n';
  }
  signed main() {
    long long q;
    cin >> q;
    while (q--) Test();
    return 0;
  }
  