#include <bits/stdc++.h>
int nextInt() {
  int x;
  scanf("%d", &x);
  return x;
}
double nextDouble() {
  double x;
  scanf("%lf", &x);
  return x;
}
long long nextLong() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
char nextChar() {
  char x;
  scanf("%c", &x);
  return x;
}
void newLine() { printf("\n"); }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long powM(long long a, long long b, long long MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > MOD) x %= MOD;
    }
    y = (y * y);
    if (y > MOD) y %= MOD;
    b /= 2;
  }
  return x;
}
long long ModExp(long long n, long long k, long long MOD) {
  long long ans = 1;
  while (k > 0) {
    if (k & 1) ans = (ans * n) % MOD;
    n = (n * n) % MOD;
    k /= 2;
  }
  return ans;
}
long long Exp(long long n, long long k) {
  long long ans = 1;
  while (k > 0) {
    if (k & 1) ans = (ans * n);
    n = (n * n);
    k /= 2;
  }
  return ans;
}
int countSetBits(long long n) {
  int ans = 0;
  while (n != 0) {
    n -= (n & -n);
    ++ans;
  }
  return ans;
}
const int N = 100100;
const int M = 100100;
const long long INF = 1e18;
const double EPS = 1e-9;
const long double PI = 2 * acos(0);
using namespace std;
const int LOG = 18;
long long mod = 1e9 + 7;
long long mod2 = 1e9 + 9;
int main() {
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  n -= 1;
  cout << n * (n + 1) * 2 + 1 << endl;
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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>
int nextInt() {
  int x;
  scanf("%d", &x);
  return x;
}
double nextDouble() {
  double x;
  scanf("%lf", &x);
  return x;
}
long long nextLong() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
char nextChar() {
  char x;
  scanf("%c", &x);
  return x;
}
void newLine() { printf("\n"); }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long powM(long long a, long long b, long long MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > MOD) x %= MOD;
    }
    y = (y * y);
    if (y > MOD) y %= MOD;
    b /= 2;
  }
  return x;
}
long long ModExp(long long n, long long k, long long MOD) {
  long long ans = 1;
  while (k > 0) {
    if (k & 1) ans = (ans * n) % MOD;
    n = (n * n) % MOD;
    k /= 2;
  }
  return ans;
}
long long Exp(long long n, long long k) {
  long long ans = 1;
  while (k > 0) {
    if (k & 1) ans = (ans * n);
    n = (n * n);
    k /= 2;
  }
  return ans;
}
int countSetBits(long long n) {
  int ans = 0;
  while (n != 0) {
    n -= (n & -n);
    ++ans;
  }
  return ans;
}
const int N = 100100;
const int M = 100100;
const long long INF = 1e18;
const double EPS = 1e-9;
const long double PI = 2 * acos(0);
using namespace std;
const int LOG = 18;
long long mod = 1e9 + 7;
long long mod2 = 1e9 + 9;
int main() {
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  n -= 1;
  cout << n * (n + 1) * 2 + 1 << endl;
  return 0;
}

```