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;
#define int ll

#define REP(i,n) for(ll i=0;i<n;++i)
#define SORT(name) sort(name.begin(), name.end())
#define ZERO(p) memset(p, 0, sizeof(p))
#define MINUS(p) memset(p, -1, sizeof(p))
#if 1
#  define DBG(fmt, ...) printf(fmt, ##__VA_ARGS__)
#else
#  define DBG(fmt, ...)
#endif

const ll LLINF = (1LL<<60);
const int INF = (1LL<<30);
const int MOD = 1000000007;
#define MAX_N 100010

// 500, 75min


ll N, X;

signed main()
{
    cin >> N >> X;

    //for(N = 2; N <= 10; ++N) {
    //    for(X = 1; X < N; ++X) {
            //printf("N: %lld X: %lld -> ", N, X);
            ll a = min(X, N - X);
            ll b = max(X, N - X);
            ll ans = 0;
            ll tmp = 0;
            if(b % a == 0) {
                ans = b + (b / a) * 2 * a;
                printf("%lld\n", ans);
                return 0;
            }
            else {
                tmp = (b / a) * 2 * a + a + b;
                ans += tmp;
                ll new_a = b % a;
                ll new_b = a;
                a = new_a; b = new_b;
            }
            while(b % a != 0) {
                tmp = (b / a) * 2 * a;
                ans += tmp;
                ll new_a = b % a;
                ll new_b = a;
                a = new_a; b = new_b;
            }
            tmp = (b / a) * 2 * a - a;
            ans += tmp;
            printf("%lld\n", ans);
    //    }
    //}

    return 0;
}