#include <bits/stdc++.h>
using namespace std;
long long int divi[129];
long long int vis[129];
map<long long int, long long int> mp;
map<long long int, char> code;
vector<int> stp[29];
string str1;
string str2;
deque<long long int> save;
int main() {
  long long int i, j, k, l = 1, p = 1, q, r, s, t, u = 4, v, w, x, y, z, n,
                         m = 2;
  cin >> n;
  if (n == 1)
    cout << "1" << '\n';
  else if (n == 2)
    cout << "5" << '\n';
  else {
    v = n * n + (n - 1) * (n - 1);
    cout << v << '\n';
  }
}

### Prompt

Please create a solution in CPP 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 divi[129];
long long int vis[129];
map<long long int, long long int> mp;
map<long long int, char> code;
vector<int> stp[29];
string str1;
string str2;
deque<long long int> save;
int main() {
  long long int i, j, k, l = 1, p = 1, q, r, s, t, u = 4, v, w, x, y, z, n,
                         m = 2;
  cin >> n;
  if (n == 1)
    cout << "1" << '\n';
  else if (n == 2)
    cout << "5" << '\n';
  else {
    v = n * n + (n - 1) * (n - 1);
    cout << v << '\n';
  }
}

```