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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

function main(input) {
  const input_array = input.split(' ');
  const sum_price = input_array[1];
  const paper_money_count = input_array[0];
  
  if(sum_price > 10000 * paper_money_count || sum_price < 1000 * paper_money_count){
    console.log('-1 -1 -1');
  }

  for(count_10000=0; count_10000<=paper_money_count ;count_10000++){
    maximum_price = 10000 * count_10000 + 5000 * (paper_money_count - count_10000);
    minimum_price = 10000 * count_10000 + 1000 * (paper_money_count - count_10000);
    if(sum_price > maximum_price || sum_price < minimum_price){
      continue;
    }
    for(count_5000=0; count_5000<=paper_money_count - count_10000 ;count_5000++){
      maximum_price = 10000 * count_10000 + 5000 * (paper_money_count - count_10000) + 1000 * (paper_money_count - count_10000 - count_5000);
      if(sum_price > maximum_price){
        continue;
      }
      for(count_1000=0; count_1000<=paper_money_count - count_10000 - count_5000 ;count_1000++){
        if(count_10000 + count_5000 + count_1000 != paper_money_count){
          continue;
        }
        if(count_10000 * 10000 + count_5000 * 5000 + count_1000 * 1000 == sum_price){
          console.log(`${count_10000} ${count_5000} ${count_1000}`);
          return;
        }
      }
    }
  }

  console.log('-1 -1 -1');
}
 
main(require('fs').readFileSync('/dev/stdin', 'utf8'));