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(ll i = 0; i < n; i++)
#define repr(i, n) for(ll i = n; i >= 0; i--)
#define inf LLONG_MAX
#define all(v) v.begin(), v.end()
using namespace std;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;


int main()
{
    ll n; cin >> n;
    ll sum = 0;
    sum += (n / 500) * 1000;
    n -= (n / 500) * 500;
    sum += (n / 5) * 5;
    cout << sum << endl;
    return 0;
}