Problem:
Score : 100 points

Problem Statement

Snuke has a favorite restaurant.

The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer.

So far, Snuke has ordered N meals at the restaurant.
Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen.
Find x-y.

Constraints

1 ≤ N ≤ 100

Input

The input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

20

Sample Output 1

15800

So far, Snuke has paid 16000 yen, and the restaurant has paid back 200 yen. Thus, the answer is 15800.

Sample Input 2

60

Sample Output 2

47200

Snuke has paid 48000 yen for 60 meals, and the restaurant has paid back 800 yen.

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

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define RFOR(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define REP(i,n) for(int i=0;i<(n);i++)
#define RREP(i,n) for(int i=(n)-1;i>=0;i--)
#define VSORT(v) sort(v.begin(), v.end())
#define DVSORT(v) sort(v.begin(), v.end(),greater<int>())
#define SORT(v, n) sort(v, v+n)
#define DSORT(v,n) sort(v, v+n,greater<int>())
#define vi vector<int>
#define pb push_back
#define P pair<int,int>


template <class T> void chmin(T&a, const T&b) { a = min(a,b); }
template <class T> void chmax(T&a, const T&b) { a = max(a,b); }

const double EPS =1e-9;
const long INF =999999999;
const long MOD =1000000007; //10^9+7
#define PI 3.14159265258979

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

int main(void){
	int N;
	cin>>N;
	printf("%d\n",800*N-200*(N/15));

	
	return 0;
}