A. Alex and a Rhombus
time limit per test: 1 second
memory limit per test: 256 megabytes
input: standard input
output:standard output

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 nn-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).

Alex asks you to compute the number of cells in a n-th order rhombus.

Input

The first and only input line contains integer nn (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 nn-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.

Analysis

This problem has a regular formula. First you count the outermost layer has how many squares. Then the second outermost layer. Until the second smallest layer.

So we add them up. Imagine we have a $n-th$ order rhombus in a cell grid. The $n-th$ rhombus has $n - 1$ layers.

So the total square except the center square is:
$$
(4n-4)+(4(n-1) - 4)+ \cdots +2 \times 4 - 4 \=4(n-1 + n - 2 + \cdots + 1) \=2 \times (n-1)n
$$
So the final answer is $2 \times (n - 1)n + 1$ .

Code

#include<bits/stdc++.h>

using namespace std;

int main() {
	int n;
	cin >> n;
	cout << 2 * n * (n - 1) + 1 << endl;
	return 0;
}

Author: o_oyao

Link: http://dyingdown.github.io/2019/11/24/CodeForces-1180A-Alex-and-a-Rhombus/

License: All articles in this blog are licensed under CC BY-NC-SA 4.0 unless stating additionally.