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>).


def factorize(x):
    res = []
    i = 2
    while i <= 5:
        while x % i == 0:
            res.append(i)
            x /= i
        i += 1
    res.append(int(x))
    return sorted(res)

def countVal(fr, x):
    for i in fr:
        if i == 2:
            x[0] += 1
        elif i == 3:
            x[1] += 1
        elif i == 5:
            x[2] += 1


a, b = map(int, input().split())
a1 = factorize(a)
b1 = factorize(b)
xa = [0, 0, 0]
xb = [0, 0, 0]
if a1[-1] > 5 or b1[-1] > 5:
    print(-1)
else:
    countVal(a1, xa)
    countVal(b1, xb)
    print(abs(xa[0]-xb[0]) + abs(xa[1]-xb[1]) + abs(xa[2]-xb[2]))
