#include <bits/stdc++.h>
using namespace std;
void merge(long long arr[], long long l, long long m, long long r) {
  long long i, j, k;
  long long n1 = m - l + 1;
  long long n2 = r - m;
  long long L[n1], R[n2];
  for (i = 0; i < n1; i++) L[i] = arr[l + i];
  for (j = 0; j < n2; j++) R[j] = arr[m + 1 + j];
  i = 0;
  j = 0;
  k = l;
  while (i < n1 && j < n2) {
    if (L[i] <= R[j]) {
      arr[k] = L[i];
      i++;
    } else {
      arr[k] = R[j];
      j++;
    }
    k++;
  }
  while (i < n1) {
    arr[k] = L[i];
    i++;
    k++;
  }
  while (j < n2) {
    arr[k] = R[j];
    j++;
    k++;
  }
}
void mergeSort(long long arr[], long long l, long long r) {
  if (l < r) {
    long long m = l + (r - l) / 2;
    mergeSort(arr, l, m);
    mergeSort(arr, m + 1, r);
    merge(arr, l, m, r);
  }
}
int main(int argc, const char* argv[]) {
  long long n;
  cin >> n;
  long long k = 1;
  for (long long j = 0; j < n; j++) {
    k = 2 * k;
  }
  cout << 2 * (k - 1);
  return 0;
}

### Prompt

Your challenge is to write 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;
void merge(long long arr[], long long l, long long m, long long r) {
  long long i, j, k;
  long long n1 = m - l + 1;
  long long n2 = r - m;
  long long L[n1], R[n2];
  for (i = 0; i < n1; i++) L[i] = arr[l + i];
  for (j = 0; j < n2; j++) R[j] = arr[m + 1 + j];
  i = 0;
  j = 0;
  k = l;
  while (i < n1 && j < n2) {
    if (L[i] <= R[j]) {
      arr[k] = L[i];
      i++;
    } else {
      arr[k] = R[j];
      j++;
    }
    k++;
  }
  while (i < n1) {
    arr[k] = L[i];
    i++;
    k++;
  }
  while (j < n2) {
    arr[k] = R[j];
    j++;
    k++;
  }
}
void mergeSort(long long arr[], long long l, long long r) {
  if (l < r) {
    long long m = l + (r - l) / 2;
    mergeSort(arr, l, m);
    mergeSort(arr, m + 1, r);
    merge(arr, l, m, r);
  }
}
int main(int argc, const char* argv[]) {
  long long n;
  cin >> n;
  long long k = 1;
  for (long long j = 0; j < n; j++) {
    k = 2 * k;
  }
  cout << 2 * (k - 1);
  return 0;
}

```