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.

Input

The first line contains two space-separated integers a and b (1 ≤ a, b ≤ 109). 

Output

If the fox is lying to the little bears and it is impossible to make the pieces equal, print -1. Otherwise, print the required minimum number of operations. If the pieces of the cheese are initially equal, the required number is 0.

Examples

Input

15 20


Output

3


Input

14 8


Output

-1


Input

6 6


Output

0
from math import gcd, sqrt
from math import log
from os import path

def factorization235(n):
    c,result = 0,1
    while n % 2 == 0:
        n /= 2; c += 1;
    while n % 3 == 0:
        n /= 3; c += 1
    while n % 5 == 0:
        n /= 5; c += 1
    if n != 1: return -1
    return c

a,b = map(int,input().split(' '))
path1,path2,div = 0,0,True
if a == b: pass
else:
    target = gcd(a,b)
    path1 = a / target; path2 = b / target
    if path1 != 1:
        path1 = factorization235(path1)
        if path1 == -1: div = False
    else: path1 = 0
    if path2 != 1:
        path2 = factorization235(path2)
        if path2 == -1: div = False
    else: path2 = 0
if div: print(int(path1 + path2))
else: print(-1)

'''
1 1024
ans:10


36 30
ans:3

20 8        GCD: 4
4  4
ans:2

20 15       GCD : 5
10 5
5  5
ans:3

If we want to reach the same number decreasing a and b by
division we have to check for GCD(a,b). GCD because our goal is also 
to minimize the number of operations, so we target the first 
common number like that.

If we are confident that both numbers will meet each other
at GCD and not earlier than we should just figure out the shortest "div" path
from both numbers to GCD and sum it up.
If it's impossible to reach GCD with 
problem div conditions from any num than return -1.

If we have GCD, we can divide a or b by it 
and check if we can the result of this operation 
divide by 2,3,5 (something like that)
[(a or b) / GCD(a,b)] / (2 or 3 or 5) = operations
if [(a or b) / GCD(a,b)] % (2 or 3 or 5) != 0: -1

If the first quotient [(a or b) / GCD(a,b)] will
be simultaneously divisible by 2 or 3 numbers from {2,3,5}
then we are making a division by the largest number. 

'''