#include <bits/stdc++.h>
using namespace std;
const long long N = 1e7;
long long B[N] = {0};
long long M = -1;
vector<long long> store;
void call(long long x, long long n, long long cnt, long long lim) {
  long long k = 2, prod = 1;
  while (prod <= lim) {
    prod = x * k;
    if (B[prod] == 1 || B[prod] == -1) {
      cnt++;
      B[prod] = -1;
      call(prod, n, cnt, lim);
      cnt--;
    }
    k++;
  }
  if (M <= cnt) M = cnt;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long A[101] = {0};
  A[1] = 1;
  A[2] = 5;
  for (long long i = 3; i <= 100; i++) {
    A[i] = 4 + A[i - 1] + A[i - 1] - A[i - 2];
  }
  long long n;
  cin >> n;
  cout << A[n] << "\n";
  return 0;
}

### Prompt

Develop a solution in CPP 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;
const long long N = 1e7;
long long B[N] = {0};
long long M = -1;
vector<long long> store;
void call(long long x, long long n, long long cnt, long long lim) {
  long long k = 2, prod = 1;
  while (prod <= lim) {
    prod = x * k;
    if (B[prod] == 1 || B[prod] == -1) {
      cnt++;
      B[prod] = -1;
      call(prod, n, cnt, lim);
      cnt--;
    }
    k++;
  }
  if (M <= cnt) M = cnt;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long A[101] = {0};
  A[1] = 1;
  A[2] = 5;
  for (long long i = 3; i <= 100; i++) {
    A[i] = 4 + A[i - 1] + A[i - 1] - A[i - 2];
  }
  long long n;
  cin >> n;
  cout << A[n] << "\n";
  return 0;
}

```