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,i,j,k,num:longint;
begin
  read(n);
  for i:=1 to 3500 do
  for j:=i to 3500 do
  begin
    if (4*i*j-n*i-n*j)<=0 then continue;
    num:=trunc((n*i*j)/(4*i*j-n*i-n*j));
  if ((n*i*j)/(4*i*j-n*i-n*j)=num) then begin writeln(i,' ',j,' ',num);exit;end;
  end;
end.