Problem:
Score : 200 points

Problem Statement

Takahashi loves gold coins. He gains 1000 happiness points for each 500-yen coin he has and gains 5 happiness points for each 5-yen coin he has. (Yen is the currency of Japan.)

Takahashi has X yen. If he exchanges his money so that he will gain the most happiness points, how many happiness points will he earn?

(We assume that there are six kinds of coins available: 500-yen, 100-yen, 50-yen, 10-yen, 5-yen, and 1-yen coins.)

Constraints

0 \leq X \leq 10^9

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

Print the maximum number of happiness points that can be earned.

Sample Input 1

1024

Sample Output 1

2020

By exchanging his money so that he gets two 500-yen coins and four 5-yen coins, he gains 2020 happiness points, which is the maximum number of happiness points that can be earned.

Sample Input 2

0

Sample Output 2

0

He is penniless - or yenless.

Sample Input 3

1000000000

Sample Output 3

2000000000

He is a billionaire - in yen.

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

#include <bits/stdc++.h>
#define FOR(i, a, b) for(int i = (a);i < (b);++i)
#define FORR(i, b, a) for (int i = (b) - 1;i >= 0;--i)
#define REP(i, n) for(int i = 0;i < (n);++i)
#define REPR(i, n) for(int i = (n) - 1;i >= 0;--i)
#define ITER(itr, v) for(auto itr = v.begin();itr != v.end();++itr)
#define SORT(v) sort(v.begin(), v.end())
#define REV(v) reverse(v.begin(), v.end())
#define SIZE(v) int(v.size())
using namespace std;
typedef long long ll;
typedef pair<int, int> P;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	ll X;
	cin >> X;
	ll res = 0;
	if (X >= 500) {
		res += X / 500 * 1000;
		X %= 500;
	}
	res += X / 5 * 5;
	cout << res << endl;

	return 0;
}