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 prime_factors(n):
    other_than_5 = False
    i = 2
    factors = []
    while i * i <= n:
        if n % i:
            i += 1
        else:
            n //= i
            if i > 5:
                other_than_5 = True
                break
            factors.append(i)
    if n > 1:
        if n > 5:
            other_than_5 = True
        factors.append(n)
    return factors, other_than_5


a, b = map(int, input().split())
factor_a, other_than_5_a = prime_factors(a)
factor_b, other_than_5_b = prime_factors(b)

if a == b:
    print(0)
elif other_than_5_a or other_than_5_b:
    print(-1)
else:
    freq_factor_a, freq_factor_b = dict(), dict()
    for dividor in [2, 3, 5]:
        freq_factor_a[dividor], freq_factor_b[dividor] = 0, 0
    for e in factor_a:
        freq_factor_a[e] += 1
    for e in factor_b:
        freq_factor_b[e] += 1
    res = sum([abs(freq_factor_a[dividor] - freq_factor_b[dividor]) for dividor in [2, 3, 5]])
    print(res)
