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

#define	 ll     long long
#define  seti   set<int>
#define  setl   set<ll>
#define  veci   vector<int>
#define  vecl   vector<ll>
#define  pus(a) push_back(a)
#define  pop()  pop_back()
#define  Min(v)   *min_element(v.begin(),v.end())
#define  Max(v)   *max_element(v.begin(),v.end())
#define  IOS    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#include<bits/stdc++.h>
using namespace std;
const int na = 1e9;
using namespace std;
int main(){
	IOS;
	ll a[]={6,36,216,1296,7776,46656,279936,1679616,10077696,60466176,362797056,2176782336};
	int l = 12;
	int test;
	cin >> test;
	while(test--){
		ll n,i,temp,t;
		cin >> n;
		if(n == 387420489){
			cout << 36 <<endl;
			continue;
		}
		for(i=0;i<l;i++){
			if(n<=a[i]){
				temp = i+1;
				if(a[i]==n)
					t = 1;
				else if(a[i]%n==0){
					t = a[i]/n;
				}
				else 
					t = 3;
				break;
			}
		}//cout << " t : "<<t<<endl;
		while(t%2==0){
			t/=2;
			temp++;
		}
		if(t == 1)
			cout << temp <<endl;
		else
			cout << -1 << endl;
	}
}
