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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
import java.math.BigInteger; 

 public class Main{
// taking inputs
static BufferedReader s1;
static BufferedWriter out;
static String read() throws IOException{String line="";while(line.length()==0){line=s1.readLine();continue;}return line;}
static int int_v (String s1){return Integer.parseInt(s1);}
static long long_v(String s1){return Long.parseLong(s1);}
static int[] int_arr() throws IOException{String[] a=read().split(" ");int[] b=new int[a.length];for(int i=0;i<a.length;i++){b[i]=int_v(a[i]);}return b;}
static long[] long_arr() throws IOException{String[] a=read().split(" ");long[] b=new long[a.length];for(int i=0;i<a.length;i++){b[i]=long_v(a[i]);}return b;}
static void assign(){s1=new BufferedReader(new InputStreamReader(System.in));out=new BufferedWriter(new OutputStreamWriter(System.out));}
// ends

//......................................@uthor_Alx..............................................


//  (Binary lifting)LCS implementation <O(NlogN,logN)>...<preprocessing,query>
//static int l,time;
//static int[][] f; // f[n][l]..
//static int[] tin, tout;
//static List<Integer>[] adj;
/*	return tin[u]<=tin[v]&&tout[u]>=tout[v]; 
}
static void dfs(int u, int p){
	tin[u]=time++;
	f[u][0]=p;
	for(int i=1;i<=l;i++){
		f[u][i]= f[f[u][i-1]][i-1];
	}

	for(int v: adj[u]){
		if(v!=p) dfs(v,u);
	}
	tout[u]=time++;
}
static int lca(int u, int v){
	if(ok(u,v)){return u;}
	if(ok(v,u)){return v;}

	for(int i=l;i>=0;i--){
		if(!ok(f[u][i],v)){
			u=f[u][i];
		}
	}
	return f[u][0];
}*/

public static void main(String[] args) throws  IOException{
	        		  assign();
              int s=int_v(read());
              long x=(long)(s/500)*1000 +((s-500*(s/500))/5)*5;
              out.write(x+"\n");
              out.flush();
	        		 
	        	}
	       
	          
	        }