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>).
a,b=map(int,input().split())
first,second=[a],[b]
dict_a,dict_b={i:[] for i in range(31)},{i:[] for i in range(31)}
dict_a[0].append(a)
dict_b[0].append(b)
c=1
for i in first:
    if i%2==0:
        first.append(i//2)
        dict_a[c].append(i//2)
    if i%3==0:
        first.append(i//3)
        dict_a[c].append(i // 3)
    if i%5==0:
        first.append(i//5)
        dict_a[c].append(i // 5)
    c+=1
c=1
for i in second:
    if i%2==0:
        second.append(i//2)
        dict_b[c].append(i//2)
    if i%3==0:
        second.append(i//3)
        dict_b[c].append(i // 3)
    if i%5==0:
        second.append(i//5)
        dict_b[c].append(i // 5)
    c+=1
f,s=set(first),set(second)
g=f.intersection(s)
if g:
    c=max(g)
    res=0
    for i in dict_a:
        if c in dict_a[i]:
            res+=i
            break
    for i in dict_b:
        if c in dict_b[i]:
            res+=i
            break
    print(res)
else:
    print("-1")
