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.

btriangle.png

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.

Input

The input is given from Standard Input in the following format:


N X


Output

Print the total length of the ray's trajectory.

Example

Input

5 2


Output

12
#!usr/bin/env python3
from collections import defaultdict
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS():return list(map(list, sys.stdin.readline().split()))
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
    l = [None for i in range(n)]
    for i in range(n):l[i] = I()
    return l
def LIR(n):
    l = [None for i in range(n)]
    for i in range(n):l[i] = LI()
    return l
def SR(n):
    l = [None for i in range(n)]
    for i in range(n):l[i] = S()
    return l
def LSR(n):
    l = [None for i in range(n)]
    for i in range(n):l[i] = SR()
    return l
mod = 1000000007

#A
"""
n = I()
l = LI()
l.sort()
ans = 0
for i in range(n):
    ans += l[2*i]
print(ans)
"""

#B
def gcd(a,b):
    if a == 0:
        return b
    return gcd(b%a,a)
n,x = LI()
print((n-gcd(n,x))*3)
#C

#D

#E

#F

#G

#H

#I

#J

#K

#L

#M

#N

#O

#P

#Q

#R

#S

#T
