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 maxn = 2e5 + 5;
  const int mod = 1e9 + 7;
  int tt;
  int n, k;
  ll der[maxn];
  ll f[maxn], dp[maxn], inv[maxn];
  ll bigmod(ll a, ll b) {
    if (b == 0) return 1ll;
    ll g = bigmod(a, b / 2);
    if (b % 2) return (((g * g) % mod) * a) % mod;
    return (g * g) % mod;
  }
  ll C(ll a, ll b) { return (((f[a] * inv[a - b]) % mod) * inv[b]) % mod; }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    f[0] = inv[0] = 1;
    for (int i = 1; i <= 2e5; i++) f[i] = (f[i - 1] * i) % mod;
    for (int i = 1; i <= 2e5; i++) inv[i] = bigmod(f[i], mod - 2);
    der[0] = 1;
    for (int i = 1; i <= 2e5; i++) der[i] = (der[i - 1] * 2) % mod;
    cin >> tt;
    while (tt--) {
      cin >> n >> k;
      for (int i = 0; i <= n; i += 2) {
        dp[i] = C(n, i);
        if (i) dp[i] = (dp[i] + dp[i - 2]) % mod;
      }
      if (n % 2) {
        ll ans = 0;
        for (int i = 0; i <= k; i++) {
          ll nw = (C(k, i) * bigmod(dp[n - 1], k - i)) % mod;
          ans = (ans + nw) % mod;
        }
        cout << ans << '\n';
      } else {
        ll ans = bigmod(dp[n - 2], k);
        for (int i = k - 1; i >= 0; i--) {
          ll nw = bigmod(dp[n - 2], k - i - 1);
          nw = (nw * bigmod(der[n], i)) % mod;
          ans = (ans + nw) % mod;
        }
        cout << ans << '\n';
      }
    }
  }
  