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++14 (GCC 5.4.1) (judge verdict: Memory Limit Exceeded)

#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <cassert>
#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <map>
#include <bitset>
using namespace std;

#define repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repl(i,0,n)
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl

#define INF 1<<30

long calc(int n, int x){
    if(n==x) return 0;
    if(n%2==0 && n/2 == x) return x*3;

    if(n/2 >=x){
        return calc(x+(n-x)%x, x) + ((n-x)/x)*x*3;
    } else {
        return calc(n, n-x);
    }
}

int main(){
    long n,x;
    cin>>n>>x;

    cout<< calc(n,x) << endl;

    return 0;
}