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: Runtime Error)

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

typedef long long ll;
#define FOR(i, a, n) for (ll i=(ll)a;i<(ll)n;i++)
#define REP(i, n) FOR(i, 0, n)

ll calc_distance(ll a, ll b){
    ll ans;

    if (a%b==0 || b%a==0){
        ans = (a>=b) ? b*(2*(a/b)-1) : a*(2*(b/a)-1);
    }
    else if (a>=b){
        ans = 2*b*(a/b) + calc_distance(a%b, b);
    }
    else{
        ans = 2*a*(b/a) + calc_distance(a, b&a);
    }
    return ans;
}

int main(){
    ll n; cin >> n;
    ll x; cin >> x;

    ll ans = n + calc_distance(n-x, x);

    cout << ans << endl;
}