description: It is the easy version of the problem. The difference is that in this version, there are no nodes with already chosen colors.
  
  Theofanis is starving, and he wants to eat his favorite food, sheftalia. However, he should first finish his homework. Can you help him with this problem?
  
  You have a perfect binary tree of 2^k - 1 nodes — a binary tree where all vertices i from 1 to 2^{k - 1} - 1 have exactly two children: vertices 2i and 2i + 1. Vertices from 2^{k - 1} to 2^k - 1 don't have any children. You want to color its vertices with the 6 Rubik's cube colors (White, Green, Red, Blue, Orange and Yellow).
  
  Let's call a coloring good when all edges connect nodes with colors that are neighboring sides in the Rubik's cube.
  
  <image>| <image>  
  ---|---  
  A picture of Rubik's cube and its 2D map.
  
  More formally: 
  
    * a white node can not be neighboring with white and yellow nodes; 
    * a yellow node can not be neighboring with white and yellow nodes; 
    * a green node can not be neighboring with green and blue nodes; 
    * a blue node can not be neighboring with green and blue nodes; 
    * a red node can not be neighboring with red and orange nodes; 
    * an orange node can not be neighboring with red and orange nodes; 
  
  
  
  You want to calculate the number of the good colorings of the binary tree. Two colorings are considered different if at least one node is colored with a different color.
  
  The answer may be too large, so output the answer modulo 10^9+7.
  
  Input
  
  The first and only line contains the integers k (1 ≤ k ≤ 60) — the number of levels in the perfect binary tree you need to color.
  
  Output
  
  Print one integer — the number of the different colorings modulo 10^9+7.
  
  Examples
  
  Input
  
  
  3
  
  
  Output
  
  
  24576
  
  
  Input
  
  
  14
  
  
  Output
  
  
  934234
  
  Note
  
  In the picture below, you can see one of the correct colorings of the first example.
  
  <image>
solution: #include <bits/stdc++.h>
  using namespace std;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const int P = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  const int N = 3e5 + 10;
  const double eps = 1e-8;
  long long power(long long a, long long b, long long p) {
    long long res = 1;
    for (; b; b >>= 1, a = a * a % p)
      if (b & 1) res = res * a % p;
    return res;
  }
  long long get_phi(long long x) {
    long long res = x, a = x;
    for (long long i = 2; i * i <= x; ++i)
      if (a % i == 0) {
        res = res / i * (i - 1);
        while (a % i == 0) a /= i;
      }
    if (a > 1) res = res / a * (a - 1);
    return res;
  }
  int main() {
    int __ = 1;
    while (__--) {
      int n;
      cin >> n;
      long long p = get_phi(P);
      long long up = (power(2, n, p) - 2 + p) % p;
      cout << 6ll * power(4, up, P) % P << "\n";
    }
    return 0;
  }
  