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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<iostream>
#include<set>
#include<stack>
#include<queue>
#include<string>
#include<algorithm>//min,max,swap,rand,reverse,sort,lower_bound
#include<functional>
#include<vector>
#include<list>
#include<map>
#include<tuple>
#include<cmath>//abs, sin, cos
#include<bitset>
using namespace std;

int main()
{
	int N, Y, t, flag=0, count=0;
	int i, j, k;
	cin >> N >> Y;
	for (i = 0; i < N; i++) {
		for (j = 0; j < N; j++) {
			for (k = 0; k < N; k++) {
				if ((10000 * k) + (5000 * j) + (1000 * i) == Y && i+j+k<=N) {
					flag = 1;
					break;
				}
			}
			if (flag == 1)
				break;
		}
		if (flag == 1)
			break;
	}
	if (flag == 1) {
		cout << k << " " << j << " " << i << endl;
	}
	else
		cout << "-1 -1 -1" << endl;

	return 0;
}