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)

// failed to generate code

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


int dp[2010][2010];
vector<int> g[2010];
int N,K;

int sdp[2010];
int size(int x,int d,int p,int tot[]){
	tot[d]++;
	for( auto e : g[x] ){
		if( e == p ) continue;
		size(e,d+1,x,tot);
	}
	return 0;
}

int tmp[2010] = {};
	
int main(){
	memset(sdp,-1,sizeof(sdp));
	for(int i = 0 ; i < 2010 ; i++)
		for(int j = 0 ; j < 2010 ; j++)
			dp[i][j] = 1e9;
	
	cin >> N >> K;
	--K;
	assert( K >= 0 );
	vector< pair<int,int> > es;
	for(int i = 0 ; i < N - 1 ; i++){
		int a,b;
		cin >> a >> b;
		--a,--b;
		g[a].push_back(b);
		g[b].push_back(a);
		es.push_back({a,b});
	}
	int ans = 1e9;
	for(int i = 0 ; i < es.size() ; i++){
		int tot1[2010] = {};
		int tot2[2010] = {};
		size(es[i].first,0,es[i].second,tot1);
		size(es[i].second,0,es[i].first,tot2);
		for(int j = 1 ; j < 2010 ; j++) tot1[j] += tot1[j-1];
		for(int j = 1 ; j < 2010 ; j++) tot2[j] += tot2[j-1];
		int sub = 1e9;
		for(int k = 0 ; k <= K ; k++){
			sub = min(N - (tot1[k] + tot2[K-k]),sub);
		}
		// cout << "[" << es[i].first << "," << es[i].second << "]" << sub << endl;
		ans = min(sub,ans);
	}
	cout << ans << endl;
	
}