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: Accepted)

#include<bits/stdc++.h>
using namespace std;

typedef long long lint;
typedef pair<int, int> P;
#define REP(i, x, n) for(lint i = x ; i < n ; i++)
#define rep(i, n) REP(i, 0, n)
#define repr(i, n) for(lint i = n - 1 ; i >= 0 ; i--)
#define SORT(c) sort((c).begin(), (c).end())

const int IINF = 1e9 + 10;
const lint LLINF = (lint)1e18 + 10;
const lint MOD = (lint)1e9 + 7;
const int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
const int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy8[] = {0, -1, -1, -1, 0, 1, 1, 1};

lint gcd(lint x, lint y){
    lint r = x % y;
    while(r != 0LL){
        x = y;
        y = r;
        r = x % y;
    }
    return y;
}

int main(){
    lint n, x;
    cin >> n >> x;
    lint ans = 3 * (n - gcd(n, x));
    cout << ans << endl;
    return 0;
}