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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#pragma GCC optimization_level 3
using ll = long long;
constexpr ll INF = 1'010'000'000'000'000'017LL;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(v) v.begin(), v.end()

int main(){
	ll n,y;
	cin>>n>>y;

	ll a,b,c;
	a=-1;
	b=-1;
	c=-1;

	for (size_t i = 0; i <= n; i++)
	{
		for (size_t j = i; j <= n; j++)
		{
			// if (n-(j-i)-i<0)
			// {
			// 	continue;
			// }
			
			if (10000*i+5000*(j-i)+1000*(n-(j-i)-i)==y)
			{
				a=i;
				b=j-i;
				c=n-i-(j-i);
			}
			
		}
		
	}
	
	cout<<a<<" "<<b<<" "<<c;
}