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>).
import sys
input = sys.stdin.readline

############ ---- Input Functions ---- ############
def inp(): # int
    return(int(input()))
def inlt(): # list
    return(list(map(int,input().split())))
def insr(): # string as char list
    s = input()
    return(list(s[:len(s) - 1]))
def instr(): # string
    return input()
def invr(): # spaced ints
    return(map(int,input().split()))

def prime_factorize(n):
    # print(n)
    factors = [0] * 3 # 2, 3, 5
    while n % 2 == 0:
        factors[0] += 1
        n //= 2
    while n % 3 == 0:
        factors[1] += 1
        n //= 3
    while n % 5 == 0:
        factors[2] += 1
        n //= 5
    # print(factors, n)
    if n > 1:
        return 0
    return factors


def get_min_ops(a, b):
    factors_a = prime_factorize(a)
    factors_b = prime_factorize(b)
    # print(factors_a, factors_b)
    if not factors_a or not factors_b:
        return -1
    num_ops = 0
    for i in range(len(factors_a)):
        num_ops += abs(factors_a[i] - factors_b[i])
    return num_ops



a, b = list(invr())
print(get_min_ops(a, b))




