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 M = 1e9 + 7;
  long long sz(long long T, long long pot) {
    if (pot == 1) {
      return T;
    }
    if (pot % 2 == 1)
      return ((sz((T * T) % M, pot / 2) * T) + M) % M;
    else
      return (sz((T * T) % M, pot / 2) + M) % M;
  }
  long long OP[200005];
  long long IV[200005];
  int main() {
    for (int i = 1; i < 200005; i++) {
      IV[i] = sz(i, M - 2);
    }
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      long long S = 0;
      long long R = 0;
      long long Q = 1;
      for (int j = 1; j <= (n - 1) / 2; j++) {
        Q *= (n - 2 * j + 2);
        Q = (Q + M) % M;
        Q *= IV[2 * j - 1];
        Q = (Q + M) % M;
        Q *= (n - 2 * j + 1);
        Q = (Q + M) % M;
        Q *= IV[2 * j];
        Q = (Q + M) % M;
        R += Q;
        R = (R + M) % M;
      }
      if (n % 2 == 0) {
        R++;
        R = (R + M) % M;
      }
      OP[0] = 1;
      for (int i = 1; i <= k; i++) {
        OP[i] = OP[i - 1] * R;
        OP[i] = (OP[i] + M) % M;
      }
      if (n % 2 == 1) {
        long long W = 1;
        long long dwa = 1;
        for (int i = 0; i <= k; i++) {
          if (i == 0) {
            if (k != 0)
              S = OP[k];
            else
              S = 1;
          } else {
            if (i != k) {
              W *= (k - i + 1);
              W = (W + M) % M;
              W *= IV[i];
              W = (W + M) % M;
              dwa *= 2;
              dwa = (dwa + M) % M;
              long long vi = dwa;
              vi *= W;
              vi = (vi + M) % M;
              vi *= OP[k - i];
              vi = (vi + M) % M;
              S += vi;
              S = (S + M) % M;
            } else {
              dwa *= 2;
              dwa = (dwa + M) % M;
              S += dwa;
            }
            S = (S + M) % M;
          }
        }
      } else {
        long long dwa = sz(2, n);
        long long Dwa = 1;
        for (int i = 0; i <= k; i++) {
          if (i == 0) {
            if (k != 0)
              S = OP[k];
            else
              S = 1;
          } else if (i == 1) {
            S += OP[k - 1];
          } else {
            Dwa *= dwa;
            Dwa = (Dwa + M) % M;
            long long vi = Dwa;
            vi *= OP[k - i];
            vi = (vi + M) % M;
            S += vi;
            S = (S + M) % M;
          }
        }
      }
      cout << S << "\n";
    }
    return 0;
  }
  