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 collections import defaultdict, Counter
from math import inf
from functools import lru_cache

import sys
#input=sys.stdin.readline


def solution():
    a,b = map(int, input().split())
    def dp(a,b):
        if a == b:
            return 0
        
        res = inf
        for i in [2,3,5]:
            if b % i == 0: 
                res = min(res, dp(a, b//i))
            if a % i == 0:
                res = min(res, dp(a//i, b))
        return res + 1
    ans = dp(a,b)
    print(ans if ans < inf else -1)

def main():
    t = 1
    #t = int(input())
    for _ in range(t):
        solution()


import sys
import threading
sys.setrecursionlimit(10**6)
threading.stack_size(1 << 27)
thread = threading.Thread(target=main)
thread.start(); thread.join()
#main()

