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>).
a,b=map(int,input().split());s=0;z,z1=[],[]
if a<b:a,b=b,a
while a%2==0 and a:
    a//=2;z+=[2]
while a%3==0 and a:
    a//=3;z+=[3]
while a%5==0 and a:
    a//=5;z+=[5]
while b%2==0 and b:
    b//=2
    if 2 in z:z.remove(2)
while b%3==0 and b:
    b//=3
    if 3 in z:z.remove(3)
while b%5==0 and b:
    b//=5
    if 5 in z:z.remove(5)
for i in z:
    if i!=2 and i!=3 and i!=5:exit(print(-1))
if a==1==b:print(len(z))
else:print(-1)