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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

int main(void) {
  int x;
  int coins[6] = {500, 5, 100, 50, 10, 1};
  int after[6];
  int point = 0;
  scanf("%d", &x);
  for(int i = 0; i < 6; i++) {
    after[i] = x / coins[i];
    x %= coins[i];
    if(i == 0) {
      point += after[i] * 1000;
    } else if(i == 1) {
      point += after[i] * 5;
    }
  }
  printf("%d", point);
  return 0;
}