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.

/*题目大意：给n个顶点、从每个顶点发出到其他各顶点的射线，且射线不相交、
 *          问这样能将这n边形分成多少个部分  n>=3
 *算法分析：给出几组样例：  顶点n	 分成部分
								3		1
								4		4
								5		9
								6		16
								7		25
								8		36
			不难发现规律	(n-4) * n + 4	(n>=4)
*/
 
#include <iostream>
#include <cstdio>
 
using namespace std;
 
int main() {
    long long int n;
    cin >> n;
    if (n == 3)
        cout << 1<< endl;
    else
        cout << (n-4) * n + 4<< endl;
 
    return 0;
}