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
# -*- coding: utf-8 -*-

import sys

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
MOD = 10 ** 9 + 7

N, X = MAP()

# 最初のN以外は、平行四辺形を小さくしていくと考える
def rec(a, b):
    a, b = min(a, b), max(a, b)
    if a == 0:
        # 最後に割り切れた時は半分でいいので減らす
        return -b
    # 同じだけ小さくする所はまとめてやる
    return a*2*(b//a) + rec(a, b%a)

print(N+rec(X, N-X))
