#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename S37, typename... Args>
void err(istream_iterator<string> it, S37 a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
const long long N = 300010, mod = 1e9 + 7, mod2 = 1e9 + 9, mod3 = 998244353,
                sq = 450, base = 727, lg = 25, inf = 1e18 + 10;
long long n, m, x, y, w, z, t, k, ans, a[N];
string s, f[N];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  x = 0;
  ans = 1;
  y = 1;
  while (y < n) {
    y++;
    ans += x + 4;
    x += 4;
  }
  cout << ans;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
void err(istream_iterator<string> it) {}
template <typename S37, typename... Args>
void err(istream_iterator<string> it, S37 a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
const long long N = 300010, mod = 1e9 + 7, mod2 = 1e9 + 9, mod3 = 998244353,
                sq = 450, base = 727, lg = 25, inf = 1e18 + 10;
long long n, m, x, y, w, z, t, k, ans, a[N];
string s, f[N];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  x = 0;
  ans = 1;
  y = 1;
  while (y < n) {
    y++;
    ans += x + 4;
    x += 4;
  }
  cout << ans;
  return 0;
}

```