#include <bits/stdc++.h>
using namespace std;
void c_p_c() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long int dp[101];
void createDP() {
  dp[0] = 0;
  dp[1] = 1;
  for (long long int i = 2; i < 101; i++) {
    dp[i] = dp[i - 1] + ((i - 1) * 4);
  }
}
int32_t main() {
  c_p_c();
  createDP();
  long long int n;
  cin >> n;
  cout << dp[n];
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
void c_p_c() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long int dp[101];
void createDP() {
  dp[0] = 0;
  dp[1] = 1;
  for (long long int i = 2; i < 101; i++) {
    dp[i] = dp[i - 1] + ((i - 1) * 4);
  }
}
int32_t main() {
  c_p_c();
  createDP();
  long long int n;
  cin >> n;
  cout << dp[n];
}

```