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: Wrong Answer)

#include <bits/stdc++.h>
#define ll long long
#define all(a) (a).begin(),(a).end()

using namespace std;

int main(int argc, char const *argv[]) {
    int N,Y;
    int a=0,b=0;
    int amari=0;
    bool cantflag=false;
    scanf("%d%d",&N,&Y);
    Y/=1000;
    const int right=Y-N;
    if(right>=9){
        a=(int)right/9;
    }
    if(right-(a*9)>=4){
        b=(int)((right-a)/4);
    }
    amari=right-(a*9)-(b*4);
    //ここでYはもう使わない
    //aが10000円、bが5000円、cが1000円の枚数

    while (1) {
        if(amari==0){
            printf("%d %d %d\n",a,b,N-(a+b));
            return 0;
        }else if(cantflag){
            printf("-1 -1 -1\n");
            return 0;
        }
        if (a>0) {
            --a;
            b+=2;
            ++amari;
            if(amari==4){
                amari-=4;
                ++b;
            }
        }else{
            cantflag=true;
        }
    }
    return 0;
}