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: Accepted)

// by Balloons
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define mpr make_pair
#define debug() puts("okkkkkkkk")
#define rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
 
using namespace std;
 
typedef long long LL;
 
const int inf = 1 << 30;
LL ans=0;
 
LL gd(LL x,LL y){
//	if(x==1 && y==1)return 2333;
	if(!y)return ans-=x,x;
	if(x<y)swap(x,y);
	ans+=x/y*y*2;
//	printf("%lld %lld %lld  %lld\n",x,y,ans,x/y*y*2);
	return gd(y,x%y);
}
 
int main(){
	LL n,x;cin>>n>>x;
	ans=n;
	gd(max(n-x,x),min(n-x,x));
	printf("%lld\n",ans);
 
	return 0;
}