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>
#define _USE_MATH_DEFINES
#include <math.h>
#include <algorithm>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define print(s) cout << s << endl
#define acc(v) accumulate(v.begin(), v.end(), 0)
#define cinv(n, v) rep(i, n) cin >> v[i]

using namespace std;
const int INF = 1e9;
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vint;
typedef vector<string> vstr;
typedef vector<char> vchar;
const ll LINF = 1e18;
const ll MOD = 1e9 + 7;
int x_pos[4] = {1, 0, -1, 0}, y_pos[4] = {0, 1, 0, -1};
int ctoi(char c)
{
	if (c >= '0' && c <= '9')
	{
		return c - '0';
	}
	return 0;
}

char upper(char c)
{
	return c - 0x20;
}

char lower(char c)
{
	return c + 0x20;
}

void unique_vector(vector<int> &v)
{
	sort(all(v));
	v.erase(unique(all(v)), v.end());
}

// n次元配列の初期化。第２引数の型のサイズごとに初期化していく。
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val)
{
	std::fill((T *)array, (T *)(array + N), val);
}

//ユークリッドの互除法
//最大公約数
ll gcd(ll x, ll y)
{
	if (y == 0)
		return x;
	return gcd(y, x % y);
}

//最小公倍数
ll lcm(ll x, ll y)
{
	ll g = gcd(x, y);
	return x / g * y;
}

ll myPow(ll x, ll n, ll m)
{
	if (n == 0)
		return 1;
	if (n % 2 == 0)
		return myPow(x * x % m, n / 2, m);
	else
		return x * myPow(x, n - 1, m) % m;
}

//k!
ll facctorialMethod(ll k)
{
	ll sum = 1;
	for (ll i = 1; i <= k; ++i)
	{
		sum *= i;
		sum %= MOD;
	}
	return sum;
}

//nCr
ll permutationMethod(ll n, ll r)
{
	ll n_sum, r_sum;
	r = n - r;
	n_sum = facctorialMethod(n);
	r_sum = facctorialMethod(r);
	return n_sum / r_sum;
}

// mod. m での a の逆元 a^{-1} を計算する
long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

ll pow_kai(int a, int n)
{ //aのn乗を計算します。
	ll x = 1;
	while (n > 0)
	{ //全てのbitが捨てられるまで。
		if (n & 1)
		{ //1番右のbitが1のとき。
			x = (x * a) % MOD;
		}
		a = (a * a) % MOD;
		n >>= 1; //bit全体を右に1つシフトして一番右を捨てる。
	}
	return x % MOD;
}

int main()
{
	int n, y;
	cin >> n >> y;
	for (int i = 0; i <= n; i++){
		for (int k = 0; k <= n; k++){
			int r = n - i - k;
			if(r < 0){
				break;
			}
			if(y == 1000*i + 5000*k + 10000*r){
				cout << r << " " << k << " " << i << endl;
				return 0;
			}
		}
	}
	cout << "-1 -1 -1" << endl;
}