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;
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
      x += 0x9e3779b97f4a7c15;
      x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
      x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
      return x ^ (x >> 31);
    }
    size_t operator()(uint64_t x) const {
      static const uint64_t FIXED_RANDOM =
          chrono::steady_clock::now().time_since_epoch().count();
      return splitmix64(x + FIXED_RANDOM);
    }
  };
  void _print() { cout << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    cout << t;
    if (sizeof...(v)) cout << ", ";
    _print(v...);
  }
  using ll = long long int;
  using ld = long double;
  template <typename T, typename V>
  using P = pair<T, V>;
  template <typename T>
  using V = vector<T>;
  constexpr ll MOD = 1e9 + 7;
  constexpr ll INF = 0x3f3f3f3f;
  constexpr ld PI = 3.1415926535897932384626;
  constexpr ld EPS = 1e-9;
  constexpr ll MAX = 1e5 + 5;
  ll power(ll a, ll b) {
    ll ans = 1;
    while (b > 0) {
      if (b % 2) (ans *= a) %= MOD;
      (a *= a) %= MOD, b >>= 1;
    }
    return ans;
  }
  void solve() {
    ll n, k;
    cin >> n >> k;
    if (n % 2) {
      ll ans = power((power(2, n - 1) + 1) % MOD, k);
      cout << ans << '\n';
    } else {
      ll ans = 0;
      for (ll x = (ll)(0); x < (ll)(k + 1); x++) {
        ll lef = power((power(2, n - 1) - 1 + MOD) % MOD, x);
        lef %= MOD;
        ll rit = power(power(2, n), (k - x - 1));
        ans += (max(lef, 1LL) * max(rit, 1LL)) % MOD;
        ans %= MOD;
      }
      cout << ans << '\n';
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    ll t = 1;
    cin >> t;
    for (ll i = (ll)(1); i < (ll)(t + 1); i++) solve();
  }
  