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 C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>

int main(void){
	long long int x;
	long long int n;
	long long int a, a_max, a_min;
	long long int b, b_max, b_min;
	long long int sum = 0;
	int vector = 0; //0:右 1:左 2:下 3:上
	scanf("%lld %lld", &n, &x);
	
	a = x;
	b = x;
	a_max = x;	
	a_min = 0;	//固定
	b_max = n;
	b_min = x;	//固定
	sum += x;
	vector = 2;
	if(x > n - x){
		x = n - x;
	}

	while(!(a == 0 && b == x)){
		
		//printf("a = %d, b = %d, vector = %d\n", a, b, vector);
		switch(vector){
			
			case 0:	//右
				b_max = b;
				sum += a_max - a;
				a = a_max;
				vector = 1;
				break;

			case 1:	//左
				if(a > b - b_min){
					vector = 2;
					sum += b - b_min;
					a = a - (b - b_min);
					b = b_min;
					
				} else {
					vector = 0;
					sum += a;
					b = b - a;
					a = 0;
					
				}
				break;
			case 2:	//下
				a_max = a;
				sum += b_max - b;
				b = b_max;
				vector = 1;
				break;
			case 3:	//上
				break;
		}
	
	}
	printf("%lld\n",sum);
	return 0;
}