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 rep(i,cc,n) for(int i=cc;i<=n;++i)
#define drep(i,cc,n) for(int i=cc;i>=n;--i)
#define sz(s) (int)(s.size())
#define vecprint(v) rep(i,0,v.size()-1)cout << v[i] << " ";cout << endl;
#define arrprint(v,n) rep(i,0,n-1)cout << v[i] << " ";cout << endl;
#define matprint(v,n,m) rep(i,0,n-1){rep(j,0,m-1){cout << v[i][j] << " ";}cout << endl;}
using namespace std;
const int mod=1000000007;
const int INF=1001001001;
typedef long long ll;
typedef pair<int,int> P;

int main(){
    int x;
    cin >> x;
    ll ans = 0;
    while(x>=500){
        ans += 1000;
        x -= 500;
    }
    while(x>=5){
        ans += 5;
        x -= 5;
    }

    cout << ans << endl;
//  printf("%.10f\n",ans);

    return 0;
}