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: Accepted)

var
  n,y,i,j,k:longint;
begin
  readln(n,y);
  if (y div n=10000) then//以下为特殊判断，会快很多
  begin
     writeln(n,' ',0,' ',0);
     halt;
  end;
  if (y div n=5000) then
  begin
    writeln(0,' ',n,' ',0);
    halt;
  end;
  if (y div n=1000) then
  begin
    writeln(0,' ',0,' ',n);
    halt;
  end;
  if (n*10000<y) then
  begin
    writeln('-1 -1 -1');
    halt;
  end;
  if (n*1000>y) then
  begin
    writeln('-1 -1 -1');
    halt;
  end;
  if (y mod 1000<>0) then
  begin
    writeln('-1 -1 -1');
    halt;
  end;//暴力开始！
  for i:=0 to y div 10000 do//从不选到至少选多少个，这样时间复杂度会少，最重要！
  for j:=0 to n-i do//n-i去重，减时间，不然过不去
  if (i*10000+j*5000+(n-i-j)*1000=y) //判断这个方案是否成立
  then begin
     writeln(i,' ',j,' ',n-i-j);//成立输出
     halt;//结束所有程序
  end;
  writeln('-1 -1 -1');//如果不成立那么他就不会结束程序，就可以输出不成立的结果拉
end.