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)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author Yulian
 */

#include <iostream>
#include <iostream>
#include <fstream>
#include <vector>
#include <set>
#include <cmath>
#include <queue>
#include <deque>
#include <string>
#include <algorithm>
#include <map>


#include <iostream>

#include <fstream>
#include <vector>
#include <set>
#include <cmath>
#include <queue>
#include <deque>
#include <string>
#include <algorithm>
#include <map>



int getTheAnswer() {
    return 42;
}

template< class InputIt>
void incAll(const InputIt &first, const InputIt &last){
  for(auto it = first; it != last; ++it){
    ++*it;
  }
}

template <class InputIt>
void printAll(const InputIt &first, const InputIt &last, std::ostream& out){
  for(InputIt it = first; it != last; ++it){
    out << *it << " ";
  }
}

// only for positive integers
int gcd(int a, int b){
  int tmp;
  if(a < b){
    tmp = a;
    a = b;
    b = tmp;
  }
  while(b != 0){
    tmp = a;
    a = b;
    b = tmp % b;
  }
  return a;
}

int lcm(int a, int b){
  return abs(a*b) / gcd(a, b);
}


using namespace std;
class TaskA {
public:
	void solve(std::istream& in, std::ostream& out) {
    ios::sync_with_stdio(false);
		int n;
		in >> n;
		n = n * 800 - (n / 15) * 200;
		out << n << endl;
	}
};


int main() {
	TaskA solver;
	std::istream& in(std::cin);
	std::ostream& out(std::cout);
	solver.solve(in, out);
	return 0;
}