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 lint = long long;
  constexpr lint maxn = 2e5 + 5;
  constexpr lint inf = 1e9;
  constexpr lint mod = 1e9 + 7;
  using namespace std;
  using pii = pair<lint, lint>;
  lint t, n, k;
  lint dp[maxn][2];
  lint exp(lint a, lint b) {
    a %= mod;
    lint ret = 1;
    while (b) {
      if (b & 1) ret *= a;
      a *= a;
      ret %= mod;
      a %= mod;
      b >>= 1;
    }
    return ret;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> t;
    while (t--) {
      cin >> n >> k;
      if (k == 0) {
        cout << 1 << '\n';
        continue;
      }
      if (n % 2 == 0) {
        lint comb = exp(2, n - 1) - 1;
        dp[0][1] = 1;
        dp[0][0] = comb;
        for (lint i = 1; i < k; i++) {
          lint prev = (dp[i - 1][0] + dp[i - 1][1]) % mod;
          dp[i][0] = (prev * comb) % mod;
          dp[i][1] = exp(exp(2, mod), n * i / mod) * exp(2, n * i % mod) % mod;
        }
      } else {
        lint comb = exp(2, n - 1);
        dp[0][1] = 1;
        dp[0][0] = comb;
        for (lint i = 1; i < k; i++) {
          lint prev = (dp[i - 1][0] + dp[i - 1][1]) % mod;
          dp[i][0] = (prev * comb) % mod;
          dp[i][1] = prev % mod;
        }
      }
      cout << (dp[k - 1][0] + dp[k - 1][1]) % mod << '\n';
    }
  }
  