Codeforces Round #328 (Div. 2)B. The Monster and the Squirrel

B. The Monster and the Squirrel

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Ari the monster always wakes up very early with the first ray of the sun and the first thing she does is feeding her squirrel.

Ari draws a regular convex polygon on the floor and numbers it's vertices 1, 2, ..., n in clockwise order. Then starting from the vertex 1 she draws a ray in the direction of each other vertex. The ray stops when it reaches a vertex or intersects with another ray drawn before. Ari repeats this process for vertex 2, 3, ..., n (in this particular order). And then she puts a walnut in each region inside the polygon.

Ada the squirrel wants to collect all the walnuts, but she is not allowed to step on the lines drawn by Ari. That means Ada have to perform a small jump if she wants to go from one region to another. Ada can jump from one region P to another region Q if and only if P and Q share a side or a corner.

Assuming that Ada starts from outside of the picture, what is the minimum number of jumps she has to perform in order to collect all the walnuts?

Input

The first and only line of the input contains a single integer n (3 ≤ n ≤ 54321) - the number of vertices of the regular polygon drawn by Ari.

Output

Print the minimum number of jumps Ada should make to collect all the walnuts. Note, that she doesn't need to leave the polygon after.

Sample test(s)

input

5

output

9

input

3

output

1

Note

One of the possible solutions for the first sample is shown on the picture above.

題目大意

       在多邊形的每個腳射出向別的腳發射光線，光線可以反射但是不能穿過已有光線，一隻松鼠在雙線間跳來跳去那松果，問要跳多少次；

解題思路：

      自己畫一下之前幾次就好；

代碼：

#include "iostream"
#include "cstdio"
#include "string.h"
#include "vector"
#include "math.h"
using namespace std;
long long n;
int main(int argc, char* argv[])
{
    cin>>n;
    cout<<(n-2)*(n-2)<<endl;
    return 0;
}