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: Wrong Answer)

/**
 * Dont raise your voice, improve your argument.
 * --Desmond Tutu
 */

#include <bits/stdc++.h>
using namespace std;

#define DEBUG
#ifndef DEBUG
#define endl "\n"
#define log(args...)
#else
#define log(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); logf(_it, args); }
void logf(istream_iterator<string> it) {
}
template<typename T, typename ... Args>
void logf(istream_iterator<string> it, T a, Args ... args) {
	cout << *it << " = " << a << endl;
	logf(++it, args...);
}
#endif

#define forn(i,k,n) for(ll i=(k); i<(n); i++)

const double PI = acos(-1);
typedef long long ll;
#define fori(n) for(ll i=0; i<(n); i++)

#define cins(s) string s; cin>>s;
#define cini(i) int i; cin>>i;
#define cinll(l) ll l; cin>>l;
#define cind(d) double d; cin>>d;
#define cinai(a, n) vi a(n); fori(n) { cin>>a[i]; }
#define cinall(a, n) vll a(n); fori(n) { cin>>a[i]; }
#define cinas(s, n) vs s(n); fori(n) { cin>>s[i]; }
#define cinad(a, n) vd a(n); fori(n) { cin>>a[i]; }

typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef vector<double> vd;
typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<string> vs;
typedef vector<ll> vll;

void solve() {
	cini(x);
	int doub = x / 500;
	int sing = x - (doub * 500);
	cout << doub * 2 + sing << endl;
}

int main() {
	cin.tie(nullptr);
	std::ios::sync_with_stdio(false);
	solve();
}