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 math import pow
def take_input(s):          #for integer inputs
    if s == 1:  return int(input())
    return map(int, input().split())

def factor(n,k):
    i = 0
    while(n%k==0):
        i += 1
        n //= k
    return i
        
a, b = take_input(2)
count = 0
if a == b:
    print(0)
    exit()

a_fac_2 = factor(a,2); a_fac_3 = factor(a,3); a_fac_5 = factor(a,5)
b_fac_2 = factor(b,2); b_fac_3 = factor(b,3); b_fac_5 = factor(b,5)
x = a
if a_fac_2>0:   x //= pow(2,a_fac_2)
if a_fac_3>0:   x //= pow(3,a_fac_3)
if a_fac_5>0:   x //= pow(5,a_fac_5)
y = b
if b_fac_2>0:   y //= pow(2,b_fac_2)
if b_fac_3>0:   y //= pow(3,b_fac_3)
if b_fac_5>0:   y //= pow(5,b_fac_5)


if x != y:
    print(-1)
else:
    print(abs(a_fac_2 - b_fac_2) + abs(a_fac_3 - b_fac_3) + abs(a_fac_5 - b_fac_5))
