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 N = 2e5 + 5;
  const int M = 2000 + 5;
  const int OO = 1e9;
  const int mod = 1e9 + 7;
  int n, k;
  long long fact[N], inv[N], cur, nxt;
  long long mul(long long a, long long b) { return (a * b) % mod; }
  long long add(long long a, long long b) { return (a + b) % mod; }
  long long sub(long long a, long long b) { return (a - b + mod) % mod; }
  long long nCr(int n, int r) {
    return ((fact[n] * inv[r]) % mod * inv[n - r]) % mod;
  }
  void calcCurAndNxt() {
    cur = nxt = 0;
    for (int i = 0; i <= n; ++i) {
      bool isOneXOR = (i & 1);
      bool isOneAND = (i == n);
      if (isOneAND == isOneXOR) {
        nxt = add(nxt, nCr(n, i));
      } else if (isOneXOR) {
        cur = add(cur, nCr(n, i));
      }
    }
  }
  long long pw(long long a, long long b) {
    if (!b) {
      return 1;
    }
    auto ret = pw(a, b / 2);
    ret = mul(ret, ret);
    if (b & 1) {
      ret = mul(ret, a);
    }
    return ret;
  }
  void calcFacAndInv() {
    fact[0] = inv[0] = 1;
    for (long long i = 1; i < N; i++) {
      fact[i] = mul(i, fact[i - 1]);
      inv[i] = pw(fact[i], mod - 2);
    }
  }
  long long solve(int bit) {
    if (!bit) {
      return 0;
    }
    return add(mul(cur, pw(2, 1LL * (bit - 1) * n)), mul(nxt, solve(bit - 1)));
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int tt;
    cin >> tt;
    calcFacAndInv();
    while (tt--) {
      cin >> n >> k;
      calcCurAndNxt();
      cout << sub(pw(2, 1LL * k * n), solve(k)) << '\n';
    }
    return 0;
  }
  