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: Time Limit Exceeded)

var
        n,y,i,j,k:longint;
begin
        readln(n,y);
        if (y div n=10000) then
        begin
                writeln(n,' ',0,' ',0);
                halt;
        end
        else if (y div n=5000) then
        begin
                writeln(0,' ',n,' ',0);
                halt;
        end
        else if (y div n=1000) then
        begin
                writeln(0,' ',0,' ',n);
                halt;
        end;
        for i:=0 to n do
            for j:=0 to n-i do
                for k:=0 to n-i-j do
                    if (i*10000+j*5000+k*1000=y)and(i+j+k=n) then
                    begin
                        writeln(i,' ',j,' ',k);
                        halt;
                    end;
        writeln(-1,' ',-1,' ',-1);
end.