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 OCaml (4.02.3) (judge verdict: Accepted)

let i_inp_list : unit -> int list = fun () ->
  let inp_split = Str.split (Str.regexp " ") (input_line stdin) in
  List.map (fun x -> int_of_string x) (inp_split)
  
let () =
  let (n, y) = Scanf.sscanf (input_line stdin) "%d %d" (fun a b -> (a, b)) in
  let solve =
    let rec loop_10000 i =
      if i > n || 10000 * i > y then None
      else
        match loop_5000 i 0 with
          None -> loop_10000 (i + 1)
        | Some x -> Some x
    and loop_5000 i j =
      if i + j > n || 10000 * i + 5000 * j > y then None
      else
        let k = (y - 10000 * i - 5000 * j) / 1000 in
        if i + j + k = n then Some (i, j, k)
        else loop_5000 i (j + 1)
    in
    match loop_10000 0 with
      None -> (-1,-1,-1)
    | Some (i,j,k) -> (i,j,k)
  in
  let (i, j, k) = solve
  in
  print_int i;
  print_char ' ';
  print_int j;
  print_char ' ';
  print_int k;
  print_newline ()