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 maxn = 400010;
  const int INF = 1e9 + 7;
  const int mod = 1e9 + 7;
  const double Pi = acos(-1.0);
  int n, m, k, t;
  long long fact[maxn], invFact[maxn];
  int dir[][2] = {{0, 0},  {0, 1},   {-1, 0}, {0, -1}, {1, 0},
                  {-1, 1}, {-1, -1}, {1, -1}, {1, 1}};
  inline int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (isdigit(ch)) {
      x = x * 10 + ch - 48;
      ch = getchar();
    }
    return x * f;
  }
  inline void print(int x) {
    if (x < 0) {
      putchar('-');
      x = -x;
    }
    if (x > 9) print(x / 10);
    putchar(x % 10 + '0');
  }
  inline int gcd(int x, int y) {
    if (!x) return y;
    return gcd(y % x, x);
  }
  inline long long power(long long x, long long k, long long p) {
    long long res = 1;
    for (; k; k >>= 1, x = (long long)x * x % p)
      if (k & 1) res = (long long)res * x % p;
    return res;
  }
  long long A(int n, int k) {
    if (k > n) {
      return 0;
    }
    return fact[n] * invFact[k] % mod * invFact[n - k] % mod;
  }
  void solve() {
    n = read();
    k = read();
    fact[0] = invFact[0] = 1;
    for (int i = 1; i <= n; ++i) {
      fact[i] = (fact[i - 1]) * i % mod;
      invFact[i] = power(fact[i], mod - 2, mod);
    }
    long long ji = 0, ou = 0;
    for (int i = 0; i <= n; ++i) {
      if (i % 2) {
      } else {
        ou = (ou + A(n, i)) % mod;
      }
    }
    long long nn = power(2, n, mod);
    long long ans = 0;
    if (n % 2 == 1)
      ans++;
    else
      ou--;
    ans = (ans + ou) % mod;
    long long now = power(ans, k, mod);
    if (n % 2 == 0) {
      for (int i = k; i >= 1; --i) {
        now += power(nn, i - 1, mod) * power(ans, k - i, mod);
        now %= mod;
      }
    }
    printf("%lld\n", now);
    return;
  }
  int main() {
    t = read();
    while (t--) {
      solve();
    }
    return 0;
  }
  