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>).
from sys import stdin,stdout
stdin.readline
def mp(): return list(map(int, stdin.readline().strip().split()))
def it():return int(stdin.readline().strip())
from collections import defaultdict as dd,Counter as C,deque
from math import ceil,gcd,sqrt,factorial,log2,floor
from bisect import bisect_right as br,bisect_left as bl
import heapq

def pf(n):
	d = dd(lambda:0)
	while n%2==0:
		d[2]+=1
		n//=2
	i=3
	while i*i<=n:
		if n%i==0:
			d[i]+=1
			n//=i
		else:
			i+=1
	if n>2:
		d[n]+=1
	return dict(d)
		
n,m = mp()

c = pf(n)
d = pf(m)

x = list(set(list(c.keys()) + list(d.keys())))

ans=0
t = set()
t.update({2,3,5})
for i in x:

	if i in c and i in d:
	# if d[i] and c[i]:
		if d[i] == c[i]:
			pass
		else:
			if i not in t:
				print(-1)
				exit()
			else:
				ans += max(c[i],d[i]) - min(c[i],d[i])
	elif i in c:
		if i not in t:
				print(-1)
				exit()
		else:
			ans += c[i]
	elif i in d:
		if i not in t:
				print(-1)
				exit()
		else:
			ans += d[i]
print(ans)

