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: Runtime Error)

#include<stdio.h>
int main()
{
    int n,m,i,num=0,c=0,sum=0;

    scanf("%d %d",&n,&m);

    int tho[n+1],wield[n+1];

    for(i=0; i<n; i++)
    {
        scanf("%d %d",&tho[i],&wield[i]);

        if(num<tho[i])num=tho[i];
    }

    for(i=0; i<n; i++)
    {
        if(num<wield[i])
        {
            sum+=wield[i];
             c++;
            if(sum>=m)
            {
                 printf("%d\n",c);
                 break;
            }
        }
    }
    //printf("sum=%d  c=%d\n",sum,c);
    if(sum<m)
    {
        sum=m-sum;

        c+=sum/num;

        if(sum%num!=0)c++;

        printf("%d\n",c);
    }

    return 0;

}