Problem:
Score : 300 points

Problem Statement

The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these.

According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.

Constraints

1 ≤ N ≤ 2000

1000 ≤ Y ≤ 2 × 10^7

N is an integer.

Y is a multiple of 1000.

Input

Input is given from Standard Input in the following format:

N Y

Output

If the total value of N bills cannot be Y yen, print -1 -1 -1.

If the total value of N bills can be Y yen, let one such set of bills be "x 10000-yen bills, y 5000-yen bills and z 1000-yen bills", and print x, y, z with spaces in between. If there are multiple possibilities, any of them may be printed.

Sample Input 1

9 45000

Sample Output 1

4 0 5

If the envelope contained 4 10000-yen bills and 5 1000-yen bills, he had 9 bills and 45000 yen in total. It is also possible that the envelope contained 9 5000-yen bills, so the output 0 9 0 is also correct.

Sample Input 2

20 196000

Sample Output 2

-1 -1 -1

When the envelope contained 20 bills in total, the total value would be 200000 yen if all the bills were 10000-yen bills, and would be at most 195000 yen otherwise, so it would never be 196000 yen.

Sample Input 3

1000 1234000

Sample Output 3

14 27 959

There are also many other possibilities.

Sample Input 4

2000 20000000

Sample Output 4

2000 0 0

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int	main(void)
{
	int x = 0;
	int y = 0;
	int z = 0;
	int i = 0;
	int j;
	int n, yen;
	scanf("%d%d",&n, &yen);
	// if (n * 1000 == yen)
	// {
	// 	printf("%d %d %d\n", x, y, n);
	// 	return (0);
	// }
	// else if (n * 5000 == yen)
	// {
	// 	printf("%d %d %d\n", x, n, z);
	// 	return (0);
	// }
	// else if (n * 10000 == yen)
	// {
	// 	printf("%d %d %d\n", n, y, z);
	// 	return (0);
	// }
	while (i <= n)
	{
		j = 0;
		while (j <= n)
		{
			if ((yen - ((10000*i) + (5000*j)) == (n-(i+j)) * 1000))
			{
				x = i;
				y = j;
				z = n - (i+j);
				break ;
			}
			j++;
		}
		i++;
	}
	if (x + y + z != n)
	{
		x = -1;
		y = -1;
		z = -1;
		printf("%d %d %d\n", x, y, z);
	}
	else if (x + y + z == n)
		printf("%d %d %d\n", x, y, z);
	return (0);
}