Two little greedy bears have found two pieces of cheese in the forest of weight *a* and *b* grams, correspondingly. The bears are so greedy that they are ready to fight for the larger piece. That's where the fox comes in and starts the dialog: "Little bears, wait a little, I want to make your pieces equal" "Come off it fox, how are you going to do that?", the curious bears asked. "It's easy", said the fox. "If the mass of a certain piece is divisible by two, then I can eat exactly a half of the piece. If the mass of a certain piece is divisible by three, then I can eat exactly two-thirds, and if the mass is divisible by five, then I can eat four-fifths. I'll eat a little here and there and make the pieces equal". 

The little bears realize that the fox's proposal contains a catch. But at the same time they realize that they can not make the two pieces equal themselves. So they agreed to her proposal, but on one condition: the fox should make the pieces equal as quickly as possible. Find the minimum number of operations the fox needs to make pieces equal.
The first line contains two space-separated integers *a* and *b* (1 ≤ *a*, *b* ≤ 10<sup class="upper-index">9</sup>).
'''input
333333333 999999999
'''
import sys
input = sys.stdin.readline
def getstr(): return input().rstrip('\r\n')
def getint(): return int(input().strip())
def getints(): return list(map(int, input().strip().split()))
def impossible(): print(-1),exit(0)

from math import *
def factorise(n, factors):
    while n % 2 == 0:
        factors.append(2)
        n //= 2
    for i in range(3,int(sqrt(n))+1,2):
        while n % i== 0:
            factors.append(i)
            n //= i
    if n > 2:
        factors.append(n)
def solve(n, factors, s, d):
	s.add(n)
	d[n] = 0
	for i in range(len(factors)):
		n//=factors[i]
		if n==1: continue
		s.add(n)
		d[n] = i+1
a,b = getints()
factors1 = []
factorise(a, factors1)
factors2 = []
factorise(b, factors2)
s1 = set()
s2 = set()
d1 = {}
d2 = {}
solve(a,factors1,s1,d1)
solve(b,factors2,s2,d2)
s = s1.intersection(s2)
if not s: impossible()
ans = 99999999999
for e in s: ans = min(ans, d1[e]+d2[e])
print(ans)