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,n) for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef long double ld;
const double pi=acos(-1.0);
const ll mod=pow(10,9)+7;
const ll INF=pow(2,31)-1;
typedef pair<int,int> P;
typedef vector<int> vi;

ll gcd(ll a,ll b){
    ll v0=a,v1=b,v2=a%b;
    while(v2!=0){
        v0=v1;
        v1=v2;
        v2=v0%v1;
    }
    return v1;
}

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

int main(){
    ios::sync_with_stdio(false);
    int x;
    cin >> x;
    ll ans=0;
    int y=x/500;
    ans+=1000*y;
    x-=500*y;
    int z=x/5;
    ans+=5*z;
    cout << ans << endl;
    return 0;
}