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

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;
import java.lang.StringBuilder;
public class Solved {

	static Scanner sc= new Scanner(System.in);
	static List<Integer> C;
	static List<Integer> B;
	
	
	
	public static void main(String[] args) {
		int t=sc.nextInt();
	//	int a;int b;
		
		
		while(t>0) {
			
			
			int n=sc.nextInt();
			int c=0;
			while(!isPowerOfTwo(n)&&n!=1) {
				
				if(n%6==0) {
					n/=6;
				}
				else n*=2;
				
				c++;
				
			}
			if(n==1)System.out.println(c);
			else System.out.println(-1);
			
				t--;
	}}
	
	
	
	
	 static boolean isPowerOfTwo (int x) 
	    { 
	      /* First x in the below expression is  
	        for the case when x is 0 */
	        return x!=0 && ((x&(x-1)) == 0); 
	          
	    } 
	
}
