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,z:int64;
    x,y:longint;
begin
  readln(n);
  for x:=1 to 3500 do
  for y:=1 to 3500 do
  if 4*x*y-(n*(x+y))>0 then
    if (x*y*n) mod (4*x*y-n*(x+y))=0 then
    begin
      z:=(x*y*n) div (4*x*y-n*(x+y));
      writeln(x,' ',y,' ',z);
      exit;
    end;
end.