Problem:
Score : 500 points

Problem Statement

Snuke is conducting an optical experiment using mirrors and his new invention, the rifle of Mysterious Light.

Three mirrors of length N are set so that they form an equilateral triangle.
Let the vertices of the triangle be a, b and c.

Inside the triangle, the rifle is placed at the point p on segment ab such that ap = X.
(The size of the rifle is negligible.)
Now, the rifle is about to fire a ray of Mysterious Light in the direction of bc.

The ray of Mysterious Light will travel in a straight line, and will be reflected by mirrors, in the same ways as "ordinary" light.
There is one major difference, though: it will be also reflected by its own trajectory as if it is a mirror!
When the ray comes back to the rifle, the ray will be absorbed.

The following image shows the ray's trajectory where N = 5 and X = 2.

It can be shown that the ray eventually comes back to the rifle and is absorbed, regardless of the values of N and X.
Find the total length of the ray's trajectory.

Constraints

2≦N≦10^{12}

1≦X≦N-1

N and X are integers.

Partial Points

300 points will be awarded for passing the test set satisfying N≦1000.

Another 200 points will be awarded for passing the test set without additional constraints.

Input

The input is given from Standard Input in the following format:

N X

Output

Print the total length of the ray's trajectory.

Sample Input 1

5 2

Sample Output 1

12

Refer to the image in the Problem Statement section.
The total length of the trajectory is 2+3+2+2+1+1+1 = 12.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {

	public static void main(String[] args) throws Exception{
		InputStreamReader is = new InputStreamReader(System.in);
		BufferedReader br = new BufferedReader (is);
		String [] input = br.readLine().split(" ");
		long N = Long.parseLong(input[0]);
		long X = Long.parseLong(input[1]);
		long result = 0;
		long NX = N-X;
		if((double)N/2 == X) {
			result = 3*X;
		}else if((double)N/2 <= X) {
			long first = NX;
			long second = N%first;
			long count = N/first;
			result = X + first*(count+1) + second * (first/(second)+1);
		}else {
			result = X + (NX) +  2*(N-NX) + (NX - (N-NX))*X+1;
		}
		System.out.println(result);
	}
}