Problem:
Score : 100 points

Problem Statement

Given an integer N not less than 3, find the sum of the interior angles of a regular polygon with N sides.

Print the answer in degrees, but do not print units.

Constraints

3 \leq N \leq 100

Input

Input is given from Standard Input in the following format:

N

Output

Print an integer representing the sum of the interior angles of a regular polygon with N sides.

Sample Input 1

3

Sample Output 1

180

The sum of the interior angles of a regular triangle is 180 degrees.

Sample Input 2

100

Sample Output 2

17640

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include<algorithm>
#include<string>
#include <cmath>
#include <vector>
#include <map>
#include <cstdio>
#include <iomanip>
#pragma region Macros
#define int long long
#define double long double
constexpr int MOD = 1000000007;
constexpr double PI = 3.14159265358979323846;
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)
#define LAST(x) x[x.size()-1]
#define ALL(x) (x).begin(),(x).end()
#define swap(a,b) (a += b,b = a - b,a -= b)
#define CEIL(a,b) ((a+b-1)/b)
int FACT(int a) {
	if (a == 0)
		return 1;
	else
		return a * FACT(a - 1);
}
int nPr(int n, int r) {
	r = n - r;
	int sum = 1;
	int i;
	for (i = n; i >= r + 1; i--)sum *= i;
	return sum;
}
//int nCr1(int n, int r)
//int nCr2(int n, int r) 
int GCD(int a, int b) // assuming a,b >= 1
{
	if (a < b)
		swap(a, b);
	if (b == 0)
		return a;
	if (a % b == 0)
		return b;
	return GCD(b, a % b);
}
int LCM(int a, int b) // assuming a,b >= 1
{
	return  a * b / GCD(a, b);
}
double LOG(int a, int b) {
	return log(b) / log(a);
}
inline bool BETWEEN(int x, int min, int max) {
	if (min <= x && x <= max)
		return true;
	else
		return false;
}

using namespace std;
#pragma endregion
signed main() {
	int n;
	cin >> n;
	cout << (n-2)*180 << endl;
}