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)

#include <stdio.h>
#include <stdlib.h>

#ifdef TEST
struct test {
    int ans;
} td[] = {
    {},
};
#endif

void
solver(long x)
{
    long n500, n5;
    n500 = x / 500;
    x -= n500 * 500;
    n5 = x / 5;
    printf("%ld\n", n500 * 1000 + n5 * 5);
}

int
main(int argc, char *argv[])
{
#ifdef TEST
    int tt;
    for(tt = 0; tt < sizeof(td) / sizeof(struct test); tt++) {
        solver();
        printf("%d\n", td[tt].ans);
    }
#endif
    long X;
    scanf("%ld", &X);
    solver(X);
    return 0;
}