Codeforces Round #569 (Div. 2) A. Alex and a Rhombus

Codeforces Round #569 (Div. 2)

 

A. Alex and a Rhombus

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).

 

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.

 

思路：感觉是找规律题，推一下找找规律就好了......

 

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<queue>
using namespace std;
#define ll long long 
const int mod=1e9+7;
const int inf=1e9+7;

//const int maxn=

int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    
    int n;
    
    while(cin>>n)
    {
        int ans=0;
        
        int f=(n<<1)-1;
        
        ans+=f;
        
        f-=2;
        
        while(f!=-1)
        {
            ans+=(f<<1);
            f-=2;
        }
        
        cout<<ans<<endl;
        
    }
    
    return 0;
}