#include <bits/stdc++.h>
using namespace std;
long long int bypass_pow(long long int a, long long int b) {
  long long int x = 1;
  a = a % (long long int)(1000000007);
  while (b > 0) {
    if (b % 2 == 1) x = (x * a) % (long long int)(1000000007);
    b /= 2;
    a = (a * a) % (long long int)(1000000007);
  }
  return x;
}
void solve(long long int n) {
  long long int t = bypass_pow((long long int)2, n);
  cout << (5 * t) % (long long int)(1000000007) << endl;
  return;
}
int main() {
  long long int n;
  cin >> n;
  long long int a = 1, ans = 1;
  if (n == 1)
    cout << n << endl;
  else {
    while (a != n) {
      ans += 4 * a;
      a++;
    }
    cout << ans << endl;
  }
  return 0;
}

### 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;
long long int bypass_pow(long long int a, long long int b) {
  long long int x = 1;
  a = a % (long long int)(1000000007);
  while (b > 0) {
    if (b % 2 == 1) x = (x * a) % (long long int)(1000000007);
    b /= 2;
    a = (a * a) % (long long int)(1000000007);
  }
  return x;
}
void solve(long long int n) {
  long long int t = bypass_pow((long long int)2, n);
  cout << (5 * t) % (long long int)(1000000007) << endl;
  return;
}
int main() {
  long long int n;
  cin >> n;
  long long int a = 1, ans = 1;
  if (n == 1)
    cout << n << endl;
  else {
    while (a != n) {
      ans += 4 * a;
      a++;
    }
    cout << ans << endl;
  }
  return 0;
}

```