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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
 
#define INF 1000000000
 
#define min(a,b) (((a)<(b))?(a):(b))
#define max(a,b) (((a)>(b))?(a):(b))

long N;
long X;
long ans = 0;

void solve(long n, long x) {
	long d = n / x;
	long m = n % x;
	ans += (d-1) * 2 * x;
	//printf("::%ld\n", (d-1) * 2 * x);
	if(m > 0) {
		solve(x+m, m);
	} else {
		return;
	}
}

int main() {
	long i, j;
	
	scanf("%ld %ld", &N, &X);
	
	if(N % 2) {
		if(N % X == 0) {
			ans = (N/X - 1) * 2 * X + N - X;
		} else {
			//ans = (N - 1) * 3;
			solve(N, X);
			ans += N;
		}
	} else {
		if(N/2 == X) {
			ans = N/2 * 3;
		} else if(N % X == 0) {
			ans = (N/X - 1) * 2 * X + N - X;
		//} else if(X % 2) {
		//	ans = (N - 1) * 3;
		} else {
			//ans = (N/2 - 1) * 6;
			solve(N, X);
			ans += N;
		}
	}
	
	printf("%ld\n", ans);
	
	return 0;
}