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)

#include <iostream>
#include <vector>
#include <string>
#include <set>
#include <list>
#include <deque>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>
#define ll long long
#define INF 0x3f3f3f3f
#define SIS std::ios::sync_with_stdio(false)
using namespace std;
ll fast_pow(ll a,ll b,ll mod)
{
    ll c=1;
    while(b>=2)
    {
        if(b%2==0)
            a*=a,b/=2;
        else
            c*=a,a*=a,b--,b/=2;
        a%=mod,c%=mod;
    }
    return (a*c)%mod;
}

int main()
{
    ll x;
    while(cin>>x)
    {
        ll a=x/500,ans=a*1000;
        x%=500;
        ans+=x%5*5;
        cout<<ans<<endl;
    }
    return 0;
}