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)

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define rep(x, a, b) for(int x = (a); x <= (b); ++x)
#define rop(x, a, b) for(int x = (a); x < (b); ++x)
#define per(x, a, b) for(int x = (a); x >= (b); --x)
using namespace std;
typedef long long LL;
typedef double DB;
LL n, x, as;
LL work(LL a, LL b) {
	if(a > b) swap(a, b);
	if(a == 0) return 0;
	return work(a, b % a) + b / a * 2 * a - a * (b % a == 0);
}
int main() {
	cin >> n >> x;
	as += n;
	as += work(x, n - x);
	cout << as << endl;
	return 0;
}