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 ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

#ifndef ONLINE_JUDGE
#define dbg(x) cerr << __LINE__ << " : " << #x << " = " << (x) << endl;
#else
#define dbg(x)
#endif

#define ff first
#define ss second

ll rec(ll A, ll B){
    if (A == B) {
        return A;
    }
    if (A == 0 || B == 0) return 0;
    ll res = 0;
    if (A < B){
        ll k = B / A;
        res += (2LL * k - 1) * A;
        ll C = B % A;
        if (C) res += A;
        res += rec(C, A);
    } else {
        ll k = A / B;
        res += (2LL * k - 1) * B;
        ll C = A % B;
        if (C) res += B;
        res += rec(C, B);
    }
    return res;
}

int main(){
    std::ios_base::sync_with_stdio(false);
    ll n, x;
    cin >> n >> x;

    ll res = n;
    res += rec(x, n - x);
    cout << res << endl;

    return 0;
}