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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
fscanf(STDIN, "%d %d", $N, $Y);
 
$k1 = $k5 = $k10 = 0;
$ans = array();
 
$tmp = $Y / 1000;
$k10max = floor($tmp / 10);
for ($i = $k10max; $i >= 0; $i--) {
    $k1 = $tmp - $i * 10;
    $k10 = $i;
 
    if ($k10 + $k1 == $N) {
        $ans[] = array("k10" => $k10, "k5" => 0, "k1" => $k1);
    } else if ($k10 + $k1 > $N) {
        $rdc = $k10 + $k1 - $N;
        if ($rdc % 4 === 0) {
            $k5 = $rdc / 4;
            $k1 -= $k5 * 5;
 
            if ($k1 >= 0) {
                $ans[] = array("k10" => $k10, "k5" => $k5, "k1" => $k1);
            }
        }
    }
}
 
if (count($ans) == 0) {
    echo "-1 -1 -1" . "\n";
} else {
    printf("%d %d %d\n", $ans[0]['k10'], $ans[0]['k5'], $ans[0]['k1']);
}