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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

var
 i,j,k,n,x,s:longint;
begin
 read(x,n);
 if x*10000<n then begin writeln('-1 -1 -1');halt;end;
 if x*1000>n then begin writeln('-1 -1 -1');halt;end;
 if n mod 1000<>0 then begin writeln('-1 -1 -1');halt;end;
 s:=x*n;
 for i:=0 to s div 10000 do
  for j:=0 to (s-i*10000) div 5000 do
   for k:=0 to (s-i*10000-j*5000) div 1000 do
    if ((i*10000+j*5000+k*1000)=s) and (i+j+k=x) then
    begin
    writeln(i,' ',j,' ',k);
    halt;
    end;
end.