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
#n=int(input())
n,m = map(int, input().strip().split(' '))
#lst = list(map(int, input().strip().split(' ')))

while(True):
    if n%2==0 and m%2==0:
        n=n//2
        m=m//2
    elif n%3==0 and m%3==0:
        n=n//3
        m=m//3
    elif n%5==0 and m%5==0:
        n=n//5
        m=m//5
    else:
        break
f=0
m1=0
m2=0
if n>m:
    n,m=m,n  #m>=n

if math.log10(n)/ math.log10(2) % 1 == 0:
    m1=2
    
elif math.log10(n)/ math.log10(3) % 1 == 0:
    m1=3
    
elif math.log10(n)/ math.log10(5) % 1 == 0:
    m1=5
    
else:
    f=1
if math.log10(m)/ math.log10(2) % 1 == 0:
    m2=2
    
elif math.log10(m)/ math.log10(3) % 1 == 0:
    m2=3

elif math.log10(m)/ math.log10(5) % 1 == 0:
    m2=5
    
else:
    f=1
if f==1:
    m2=0
    if m%n==0:
        k3=m//n
        if math.log10(k3)/ math.log10(2) % 1 == 0:
            m2=2

        elif math.log10(k3)/ math.log10(3) % 1 == 0:
            m2=3

        elif math.log10(k3)/ math.log10(5) % 1 == 0:
            m2=5
        else:
            f=2
        if f==2:
            print(-1)
        else:
            print(int(math.log10(k3)/ math.log10(m2)))
    else:
        print(-1)
else:
    k1=math.log10(n)/ math.log10(m1)
    k2=math.log10(m)/ math.log10(m2)
    print(int(k1)+int(k2))
    
