#include <bits/stdc++.h>
using namespace std;
bool Check(unsigned int N, int pos) { return (bool)(N & (1LL << pos)); }
long long Set(unsigned int N, int pos) { return N = N | (1LL << pos); }
int dx[8] = {0, 0, 1, -1, -1, 1, 1, -1};
int dy[8] = {1, -1, 0, 0, 1, -1, 1, -1};
bool chk(int x, int pos) { return (bool)(x & (1 << pos)); }
int Set(int x, int pos) { return x = x | (1 << pos); }
int clr(int x, int pos) { return x = x & ~(1 << pos); }
int n, m, a[100005];
int main() {
  int ii, i, j, k, z = 0, c = 0, t, x, y, l, r, mid;
  scanf("%d", &n);
  cout << n * n + (n - 1) * (n - 1) << endl;
}

### 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;
bool Check(unsigned int N, int pos) { return (bool)(N & (1LL << pos)); }
long long Set(unsigned int N, int pos) { return N = N | (1LL << pos); }
int dx[8] = {0, 0, 1, -1, -1, 1, 1, -1};
int dy[8] = {1, -1, 0, 0, 1, -1, 1, -1};
bool chk(int x, int pos) { return (bool)(x & (1 << pos)); }
int Set(int x, int pos) { return x = x | (1 << pos); }
int clr(int x, int pos) { return x = x & ~(1 << pos); }
int n, m, a[100005];
int main() {
  int ii, i, j, k, z = 0, c = 0, t, x, y, l, r, mid;
  scanf("%d", &n);
  cout << n * n + (n - 1) * (n - 1) << endl;
}

```