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 math

numbers = list(map(int , input().split()))
data_results = {}
min_step     =0
def search_for_commen_factor (number , steps , commen_factor):
    arr   = [2,3,5]
    if commen_factor == number  :
        return  [True , steps]
    for i in arr :
        if number % i == 0 :
            if (number / i) >= commen_factor :
                x = search_for_commen_factor( math.ceil(number / i) , steps+1 , commen_factor)
                if x[0] == True :
                    return x ;
    return [False , steps]
def highest_common_factor(a, b):
    while b != 0:
        a, b = b, a % b
    return a
commen_factor = highest_common_factor(numbers[0] , numbers[1])

check = 0
for i in numbers :
    x =  search_for_commen_factor(i, 0, commen_factor)
    if x[0] == True :
        check = 1
        min_step += x[1]
    else :
        check = 0
if check == 0 :
    print(-1)
else :
    print(min_step)