#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
int dx[8] = {1, -1, 0, 0, 1, 1, -1, -1};
int dy[8] = {0, 0, 1, -1, 1, -1, 1, -1};
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  cout << binpow(2, n + 1) - 2;
}

### Prompt

Your task is to create a CPP solution to the following problem:

The numbers of all offices in the new building of the Tax Office of IT City will have lucky numbers.

Lucky number is a number that consists of digits 7 and 8 only. Find the maximum number of offices in the new building of the Tax Office given that a door-plate can hold a number not longer than n digits.

Input

The only line of input contains one integer n (1 ≤ n ≤ 55) — the maximum length of a number that a door-plate can hold.

Output

Output one integer — the maximum number of offices, than can have unique lucky numbers not longer than n digits.

Examples

Input

2


Output

6

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
int dx[8] = {1, -1, 0, 0, 1, 1, -1, -1};
int dy[8] = {0, 0, 1, -1, 1, -1, 1, -1};
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  cout << binpow(2, n + 1) - 2;
}

```