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.*;

import javax.lang.model.util.ElementScanner6;
public class snippe{
    static void print(String s){
        System.out.println(s);
    }
    static void print(int  s){
        System.out.println(s);
    }
    static int[] rep(int[] arr,int n){
        Scanner sc=new Scanner(System.in);
        for(int i=0;i<n;i++){
            arr[i]=sc.nextInt();
        }
        return arr;
    }
    static void solve(){
        Scanner sc=new Scanner(System.in);
        // int n=sc.nextInt();
      



    }
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
     
    // method to return LCM of two numbers
    static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }
    static long spd(long n){
        for(int i=2;i*i<=n;i++){
            if(n%i==0){
                return i;
            }
        }
        return n;


    }
    public static int[] primeFactors(long n)
    {
        // Print the number of 2s that divide n
        int[]arr=new int[3];
        while (n%2==0)
        {
            arr[0]++;
            n /= 2;
        }
 
        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(n); i+= 2)
        {
            // While i divides n, print i and divide n
            while (n%i == 0)
            {   if(i==3){
                arr[1]++;}
                else{
                    arr[2]=1;
                }
                n /= i;
            }
        }
 
        // This condition is to handle the case whien
        // n is a prime number greater than 2
        if (n>2){
            int[]a=new int[3];
            a[2]=1;
            return a;
        }
            

            
            return arr;
    }




     
    public static void main(String[] args) {
    Scanner sc=new Scanner(System.in);
        int k=sc.nextInt();
        while(k-->0){
            int n=sc.nextInt();
            	int cnt2 = 0, cnt3 = 0;
		while (n % 2 == 0) {
			n /= 2;
			++cnt2;
		}
		while (n % 3 == 0) {
			n /= 3;
			++cnt3;
		}
		if (n == 1 && cnt2 <= cnt3) {
			System.out.println(2*cnt3-cnt2);
		} else {
			System.out.println(-1);
		}
            
           
















        }
    
    
 

       






     

        
}



    

}