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>).
cntOne,cntTwo = list(map(int,input().split()))
result = 0
arrDel = []
arrBel = []
c = 0
c1 = 0
pn = 0
if cntOne > cntTwo:
    pn = cntOne
else:
    pn = cntTwo
for i in range(pn):
    if cntOne == 1 and cntTwo == 1:
        break
    if cntOne < cntTwo or c == 1:
        if cntTwo % 2 == 0:
            cntTwo //= 2
            arrDel.append(2)
        elif cntTwo % 3 == 0:
            cntTwo //= 3
            arrDel.append(3)
        elif cntTwo % 5 == 0:
            cntTwo //= 5
            arrDel.append(5)
    else:
        if cntOne % 2 == 0:
            cntOne //= 2
            arrBel.append(2)
        elif cntOne % 3 == 0:
            cntOne //= 3
            arrBel.append(3)
        elif cntOne % 5 == 0:
            cntOne //= 5
            arrBel.append(5)
        else:
            c += 1
result = result + abs(arrDel.count(2) - arrBel.count(2))
result = result + abs(arrDel.count(3) - arrBel.count(3))
result = result + abs(arrDel.count(5) - arrBel.count(5))
if cntTwo != cntOne:
    print('-1')
else:
    print(result)