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：感受是找規律題，推一下找找規律就行了......

 

 

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<algorithm>
 6 #include<map>
 7 #include<set>
 8 #include<vector>
 9 #include<queue>
10 using namespace std; 11 #define ll long long 
12 const int mod=1e9+7; 13 const int inf=1e9+7; 14 
15 //const int maxn=
16 
17 int main() 18 { 19     ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 20     
21     int n; 22     
23     while(cin>>n) 24  { 25         int ans=0; 26         
27         int f=(n<<1)-1; 28         
29         ans+=f; 30         
31         f-=2; 32         
33         while(f!=-1) 34  { 35             ans+=(f<<1); 36             f-=2; 37  } 38         
39         cout<<ans<<endl; 40         
41  } 42     
43     return 0; 44 }