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>).
#include <bits/stdc++.h> 
#include <iostream>
#include<vector>
#define int long long

using namespace std;

vector<int> primeFactors(int n)
{
    vector<int> prime;
    if (n%2==0)
    {
        prime.push_back(2);
        while(n%2==0) n = n / 2;
    }
    for (int i=3;i<=sqrt(n);i=i+2)
    {
        if (n%i==0)
        {
            prime.push_back(i);
            while(n%i==0) n/=i;
        }
    }
    if (n>2) prime.push_back(n);
    return prime;

}

int getPower(int n, int i)
{
    int ans = 0;
    while(n%i==0) 
    {
        n/=i;
        ans++;
    }
    return ans;
}
signed main()
{
   int a,b;
   cin>>a>>b;
   vector<int> pa = primeFactors(a);
   vector<int> pb = primeFactors(b);
   int f = 0;
   for (auto i:pa)
   {
       if (i!=2 or i!=3 or i!=5) f = 1;
   }
   for (auto i:pb)
   {
       if (i!=2 or i!=3 or i!=5) f = 1;
   }
   if (f==7)
   {
       cout<<-1;
   }
   else
   {
       int aa[5];
       int tmp=0;
       int ans = 0;
       for (auto i:pb)
       {
           
           if (i==2 or i==3 or i
