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>
using namespace std;
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<long>;
using vs = vector<string>;
using vvi = vector<vector<int>>;
using vvb = vector<vector<bool>>;
using vvc = vector<vector<char>>;
using pii = pair<int, int>;
using pll = pair<long, long>;
#define fix20 cout << fixed << setprecision(20);
#define YES cout << "Yes" << endl;
#define NO cout << "No" << endl;
#define rep(i,n) for(int i=0; i<(int)(n);i++)
#define rrep(i,n) for(int i=(n)-1;i>=0;i--)
#define MOD 1000000007
#define all(vec) vec.begin(), vec.end()

int main(){
    long x;
    cin >> x;
    long ans = (x/500)*1000;
    long res = x%500;
    cout << ans + (res/5)*5 << endl;
}