#include <bits/stdc++.h>
using namespace std;
vector<long long int> dp(101);
void compute() {
  dp[1] = 1;
  long long int d = 4;
  for (long long int i = 2; i <= 100; i++) {
    dp[i] = dp[i - 1] + d;
    d += 4;
  }
}
void solve() {
  long long int t, n, m, x, y, z, d, p = 0, b = 0, l = INT_MAX, k, mini = 0,
                                     prev;
  compute();
  cin >> n;
  cout << dp[n];
}
int main() {
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### 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;
vector<long long int> dp(101);
void compute() {
  dp[1] = 1;
  long long int d = 4;
  for (long long int i = 2; i <= 100; i++) {
    dp[i] = dp[i - 1] + d;
    d += 4;
  }
}
void solve() {
  long long int t, n, m, x, y, z, d, p = 0, b = 0, l = INT_MAX, k, mini = 0,
                                     prev;
  compute();
  cin >> n;
  cout << dp[n];
}
int main() {
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```