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 (GCC 5.4.1) (judge verdict: Accepted)

/*In the name of Allah*/
#include<stdio.h>
#include<math.h>
#include<limits.h>
#include<assert.h>
#include<stdbool.h>
#include<stddef.h>
#include<stdint.h>
#include<string.h>
#include<unistd.h>
#include<stdlib.h>
#include<ctype.h>
#include<time.h>
typedef long long ll;
typedef long double ld;

#define s(a) scanf("%lld",&a)
#define p(a) printf("%lld",a)
#define nl printf("\n")
#define space printf(" ")
#define pi acos(-1)

#define FOR1(i,n) for(ll i=1;i<=n;i++)
#define FOR0(i,n) for(ll i=0;i<n;i++)

int main()
{ 
ll n,t,sum=0;
    s(n);
     t=n/500; n%=500;t=t*1000;
    sum+=t; t=n/5; t=t*5;
    sum+=t;
    
    p(sum);
    
    
    nl; 
  



return 0;
}