Problem:
Score : 300 points

Problem Statement

You are given an integer N.

Find a triple of positive integers h, n and w such that 4/N = 1/h + 1/n + 1/w.

If there are multiple solutions, any of them will be accepted.

Constraints

It is guaranteed that, for the given integer N, there exists a solution such that h,n,w \leq 3500.

Inputs

Input is given from Standard Input in the following format:

N

Outputs

Print a triple of positive integers h, n and w that satisfies the condition, in the following format:

h n w

Sample Input 1

2

Sample Output 1

1 2 2

4/2 = 1/1 + 1/2 + 1/2.

Sample Input 2

3485

Sample Output 2

872 1012974 1539173474040

It is allowed to use an integer exceeding 3500 in a solution.

Sample Input 3

4664

Sample Output 3

3498 3498 3498

Solution in Pascal (FPC 2.6.2) (judge verdict: Accepted)

var
	N,B,C:int64;
	A:Longint;
begin
	read(N);
	if N mod 2=0 then begin
		writeln(N,' ',N,' ',N div 2);
		exit;
	end else if N mod 3=0 then begin
		writeln(N*2,' ',N*2,' ',N div 3);
		exit;
	end;
	for A:=3*N div 4 downto 1 do begin
		B:=2*N*A div(4*A-N);
		while 4*A*B>N*(A+B)do begin
			if N*A*B mod(4*A*B-N*(A+B))=0 then begin
				writeln(A,' ',B,' ',N*A*B div(4*A*B-N*(A+B)));
				exit;
			end;
			dec(B);
		end;
	end;
end.