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 collections import Counter
def solve(n):
    t = []
    while n % 2 == 0 :
        t.append(2)
        n//= 2
        
    i = 3 
    while i * i <= n :
        while n % i == 0 :
            t.append(i)
            n//= i
        i += 1

    if n > 2 :
        t.append(n)
    return t

a , b = map(int,input().split())
if a == b :print(0);exit()

#print(solve(a))
#print(solve(b))
r1 = Counter(solve(a))
r2 = Counter(solve(b))
#print(r1)
#print(r2)

cnt1 , cnt2 = 0 , 0
flag = True
for i in r1.keys() :
    if i in r2.keys() and r1[i] != r2[i]:
        if (i % 2 == 0 or i % 3 == 0 or i % 5 == 0 ):
            cnt1 += abs(r1[i] - r2[i])
        else:
            flag = False
            break

    elif i not in r2.keys():
        if (i % 2 == 0 or i % 3 == 0 or i % 5 == 0 ):
            cnt1+= r1[i]
        else:
            flag = False
            break

if flag == False:
    print('-1')
    exit(0)

for i in r2.keys() :
    if i not in r1.keys() :
        if (i % 2 == 0 or i % 3 == 0 or i % 5 == 0 ):
            cnt2 += r2[i]
        else:
            flag = False
            break
if flag == False:
    print(-1)
    exit(0)

print(cnt1 + cnt2)


