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>
#include <math.h>
#include <string.h>
#include <stdlib.h>


int main(void)
{
	//変数の宣言
	long long int n,x;
	long long int a,b;
	
	//よく使う変数
	int i,j,k,l;
	int flag=0;
	long long int ans=0;
	int count=0;
	int temp,temp1,temp2;
	int min;
	//データの読み込み
	scanf("%lld %lld",&n,&x);
	
	/*for(i=0;i<n;i++){
		for(j=0;j<n;j++){
			scanf("%d",&a[i][j]);
		}
	}*/
	
//	printf("nは%dです\n", n);
//	printf("データの読み込み終了\n");
	//実際の処理
	
	ans=n;
	a=x;
	b=n-x;
	
	while(flag==0){
		if(a==b){
			ans=ans+a;
			flag=1;
		}else if(a<b){
			ans=ans+a*2;
			b=b-a;
		}else{
			ans=ans+b*2;
			a=a-b;
		}
	}
	
//	printf("計算部分終了\n");
	//出力
	
	printf("%lld",ans);
	
//	printf("結果の出力終了\n");
	
	return 0;
	
}