#include <bits/stdc++.h>
using namespace std;
long long int mod = 1e9 + 7;
long long int power(long long int num, long long int r) {
  if (r == 0) return 1;
  if (r == 1) return num % mod;
  long long int ans = power(num, r / 2) % mod;
  if (r % 2 == 0) {
    return (ans * ans) % mod;
  }
  return (((ans * ans) % mod) * num) % mod;
}
long long int modinv(long long int num) { return power(num, mod - 2); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  if (n == 1) {
    cout << "1";
    return 0;
  }
  cout << 2 * n * n - 2 * n + 1;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.

A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).

A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).

<image>

Alex asks you to compute the number of cells in a n-th order rhombus.

Input

The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.

Output

Print exactly one integer — the number of cells in a n-th order rhombus.

Examples

Input


1


Output


1

Input


2


Output


5

Input


3


Output


13

Note

Images of rhombus corresponding to the examples are given in the statement.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int mod = 1e9 + 7;
long long int power(long long int num, long long int r) {
  if (r == 0) return 1;
  if (r == 1) return num % mod;
  long long int ans = power(num, r / 2) % mod;
  if (r % 2 == 0) {
    return (ans * ans) % mod;
  }
  return (((ans * ans) % mod) * num) % mod;
}
long long int modinv(long long int num) { return power(num, mod - 2); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  if (n == 1) {
    cout << "1";
    return 0;
  }
  cout << 2 * n * n - 2 * n + 1;
}

```