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 <bits/stdc++.h>
#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)  
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define ll long long
#define pb(a) push_back(a)
#define INF 999999999
#define MOD 1000000007
using namespace std;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> PP;
typedef pair<ll, LP> LPP;

int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};

int gcd(int a,int b){return b?gcd(b,a%b):a;}

ll modpow(ll n,ll k){
  if(k == 1){
    return n;
  }else if(k == 0){
    return 1;
  }else if(k%2==0){
    ll memo = modpow(n,k/2)%MOD;
    return memo*memo%MOD;
  }else{
    return n * modpow(n,k-1)%MOD;
  }
}

int main(){

  int n;
  cin>>n;
  cout<<(n-2)*180<<endl;

 

  return 0;
}