You are given an integer n. In one move, you can either multiply n by two or divide n by 6 (if it is divisible by 6 without the remainder).

Your task is to find the minimum number of moves needed to obtain 1 from n or determine if it's impossible to do that.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow. 

The only line of the test case contains one integer n (1 ≤ n ≤ 10^9).

Output

For each test case, print the answer — the minimum number of moves needed to obtain 1 from n if it's possible to do that or -1 if it's impossible to obtain 1 from n.

Example

Input


7
1
2
3
12
12345
15116544
387420489


Output


0
-1
2
-1
-1
12
36

Note

Consider the sixth test case of the example. The answer can be obtained by the following sequence of moves from the given integer 15116544:

  1. Divide by 6 and get 2519424; 
  2. divide by 6 and get 419904; 
  3. divide by 6 and get 69984; 
  4. divide by 6 and get 11664; 
  5. multiply by 2 and get 23328; 
  6. divide by 6 and get 3888; 
  7. divide by 6 and get 648; 
  8. divide by 6 and get 108; 
  9. multiply by 2 and get 216; 
  10. divide by 6 and get 36; 
  11. divide by 6 and get 6; 
  12. divide by 6 and get 1. 
1
999838675

-1

/// In the name of Almighty Allah
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define el  '\n'
#define lcm(a, b) a/__gcd(a, b) * b
#define all(x) x.begin(), x.end()
#define ints(x) x-'0'
#define IOS ios_base::sync_with_stdio(false); cin.tie(0);cout.tie(0);
int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};
int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};
ll Pow(ll a, ll x){ll res = 1;while (x > 0){if (x % 2 != 0)res = (res * a) ;a = (a * a) ;x >>= 1;}return res;}
template <class... Args>auto &see(const Args &... args) { return ((cerr << " " << args), ...) << '\n'; }
#define bug(...) cerr << "[ " #__VA_ARGS__ " ]=";see(__VA_ARGS__) << '\n';



int t;

int main()
{

    IOS
     cin >> t ;

     while(t--)
     {
         int  n ;

         cin >> n ;

         int mi = 0 ;

         while(n>1)
         {
             if(n%6==0) n /= 6 ;

             else n *= 2 ;

             mi++;
         }
         if(n != 1) mi= -1 ;
         cout << mi << el ;
    }

}
