#include <bits/stdc++.h>
using namespace std;
static int rowNbr[] = {-1, -1, -1, 0, 0, 1, 1, 1};
static int colNbr[] = {-1, 0, 1, -1, 1, -1, 0, 1};
long long int power(long long int base, long long int p) {
  long long int ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base);
    base = (base * base);
    p /= 2;
  }
  return ans;
}
long long int modPower(long long int base, long long int p,
                       long long int mod = (1000000007LL)) {
  long long int ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
set<long long int> X;
set<long long int> Y;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  cout << power(2, 1 + n) - 2;
}

### Prompt

Create a solution in CPP for 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;
static int rowNbr[] = {-1, -1, -1, 0, 0, 1, 1, 1};
static int colNbr[] = {-1, 0, 1, -1, 1, -1, 0, 1};
long long int power(long long int base, long long int p) {
  long long int ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base);
    base = (base * base);
    p /= 2;
  }
  return ans;
}
long long int modPower(long long int base, long long int p,
                       long long int mod = (1000000007LL)) {
  long long int ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
set<long long int> X;
set<long long int> Y;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  cout << power(2, 1 + n) - 2;
}

```