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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<tuple>
#include<cmath>
#include<iomanip>


#define MOD 1000000007
#define INF 1001001001
#define MIN -1001001001
#define rep(i,k,N) for(int i=k;i<N;i++)
#define MP make_pair
#define MT make_tuple //tie,make_tuple は別物
#define PB push_back
#define PF push_front
#define all(x) (x).begin(),(x).end()


using namespace std;
typedef long long ll;
typedef vector<ll> v;
typedef vector<vector<ll>> vv;

int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

int main(){
    ll X;
    cin>>X;
    ll ans = 0;
    ans += X/500;
    ans *= 1000;
    ll n = (X%500)/5;
    n *= 5;
    ans += n;
    cout<<ans;
    return 0;
}