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<iostream>
#include<list>
#include<string>
#include<cstring>
#include<sstream>
#include<cctype>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<stack>
#include<fstream>
#include<cstdlib>
#include<vector>
#include<map>
#include<set>
#include<utility>
#include<iomanip>
#include<queue>
using namespace std;

#define LL long long int
#define uLL unsigned long long int

#define S(a) scanf("%d",&a)
#define S2(a,b) scanf("%d%d",&a,&b)
#define S3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define SLL(a) scanf("%lld",&a)
#define SLL2(a,b) scanf("%lld%lld",&a,&b)
#define SLL3(a,b,c) scanf("%lld%lld%lld",&a,&b,&c)
#define SC(a) scanf("%c",&a)
#define P(a) printf("%d",a)
#define PS(a) printf("%s",a)
#define PLL(a) printf("%lld",a)
#define PCASE(kk) printf("Case %d: ",kk++)
#define PCASENL(kk) printf("Case %d:\n",kk++)
#define NL puts("")

#define pb push_back
#define mp(a,b) make_pair(a,b)
#define pi (2.0*acos(0.0))
#define pii pair<int,int>

template<typename T>inline T POW(T B,T P)
{
    if(P==0) return 1;
    if(P&1) return B*POW(B,P-1);
    else return SQR(POW(B,P/2));
}
template <typename T>inline T ModInv (T b,T m)
{
    return BigMod(b,m-2,m);
}
template<typename T>inline T ABS(T a)
{
    if(a<0)return -a;
    else return a;
}
template<typename T>inline T gcd(T a,T b)
{
    if(a<0)return gcd(-a,b);
    if(b<0)return gcd(a,-b);
    return (b==0)?a:gcd(b,a%b);
}
template<typename T>inline T lcm(T a,T b)
{
    if(a<0)return lcm(-a,b);
    if(b<0)return lcm(a,-b);
    return a*(b/gcd(a,b));
}
template <class T> inline T BMOD(T p,T e,T m)
{
    T ret=1;
    while(e)
    {
        if(e&1) ret=(ret*p)%m;
        p=(p*p)%m;
        e>>=1;
    }
    return (T)ret;
}

//for(__typeof(pp.begin()) i=pp.begin(); i!=pp.end(); i++ )

//knight and king move....



//int Dx[]={-2,-1,1,2,1,2,-2,-1};
//int Dy[]={-1,-2,2,1,-2,-1,1,2};
//int dx[]={-1,1,0,0};
//int dy[]={0,0,-1,1};
//int dx[]= {-1,-1,0,0,1,1};
//int dy[]= {-1,0,-1,1,0,1};
//////////////////////////////////////////////////

int main()
{
    LL n,x;
    SLL2(n,x);

    LL ds=(n-(2*x));

    if(ds==0)
    {
        PLL((x*3LL)),NL;
    }
    else if(ds>0)
    {
        LL ans;

        ans=3LL*x+(n-x);
        LL l=x,w=ds;
        if(l<w)swap(l,w);
        while(1)
        {
            if(l%w==0)
            {
                ans+=((2LL*(l/w))-1)*w;
                break;
            }
            ans+=(2LL*(l/w))*w;
//            l=l-((2LL*(l/w))*w);
            l%=w;
            if(l==0)break;
            if(l<w)swap(l,w);

        }
        PLL(ans),NL;
    }
    else
    {
        ds=n-x;
        LL ans=x+ds;
        LL l=x,w=ds;
        if(l<w)swap(l,w);
        while(1)
        {
            if(l%w==0)
            {
                ans+=((2LL*(l/w))-1)*w;
                break;
            }
            ans+=(2LL*(l/w))*w;
//            l=l-((2LL*(l/w))*w);
            l%=w;
            if(l==0)break;
            if(l<w)swap(l,w);

        }
        PLL(ans),NL;
    }



    return 0;
}