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;
  using ll = long long;
  const int MONKE = 0;
  ll inv_mod(ll a, ll m) {
    ll _m = m, q, t, x, y;
    if (m == 1) return 0;
    for (y = 0, x = 1; a > 1; t = y, y = x - q * y, x = t) {
      q = a / m, t = m;
      m = a % m, a = t;
    }
    return x < 0 ? x + _m : x;
  }
  class modulo_int {
   public:
    ll val;
    static const ll mod = 1e9 + 7;
    modulo_int(ll _val = 0) { val = _val > 0 ? _val % mod : _val % mod + mod; }
    modulo_int operator+(modulo_int rhs) { return modulo_int((val + rhs.val)); }
    modulo_int operator-(modulo_int rhs) { return modulo_int((val - rhs.val)); }
    modulo_int operator*(modulo_int rhs) { return modulo_int((val * rhs.val)); }
    modulo_int operator/(modulo_int rhs) {
      return modulo_int(binpow(rhs, mod - 2) * val);
    }
    void operator+=(modulo_int rhs) { *this = *this + rhs; }
    void operator-=(modulo_int rhs) { *this = *this - rhs; }
    void operator*=(modulo_int rhs) { *this = *this * rhs; }
    void operator/=(modulo_int rhs) { *this = *this / rhs; }
    friend modulo_int binpow(modulo_int val, ll p) {
      modulo_int ans = 1;
      for (; p > 0; p >>= 1) {
        if (p & 1) ans = ans * val;
        val *= val;
      }
      return ans;
    }
    friend ostream& operator<<(ostream& o, modulo_int& a) {
      o << a.val;
      return o;
    }
    friend istream& operator>>(istream& o, modulo_int& a) {
      o >> a.val;
      return o;
    }
  };
  void solve() {
    int n, k;
    cin >> n >> k;
    vector<modulo_int> DP(k + 1);
    DP[0] = 1;
    modulo_int two = 2;
    modulo_int ways = binpow(two, n - 1);
    for (int i = 1; i <= k; i++) {
      if (n & 1)
        DP[i] = DP[i - 1] * (ways + 1);
      else
        DP[i] = DP[i - 1] * (ways - 1) + binpow(binpow(two, i - 1), n);
    }
    cout << DP[k] << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t = 1;
    cin >> t;
    for (int tc = 1; tc <= t; tc++) {
      solve();
    }
    return MONKE;
  }
  