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 sys
import bisect as b
import math
from collections import defaultdict as dd
input=sys.stdin.readline
#sys.setrecursionlimit(10**7)
def cin():
    return map(int,sin().split())
def ain():            
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
for _ in range(inin()):
    n=inin()
    p2=0
    p3=0
    while(n%2==0):
        p2+=1
        n=n//2
    while(n%3==0):
        p3+=1
        n//=3
    if(n==1):
        if(p3>=p2):
            if(p3==p2):
                print(p3)
            elif(p2==0):
                print(p3*2)
            else:
                print(p3+p3-p2)
        else:
            print(-1)

    else:
        print(-1)
    
    
    
    
        
                
        
        
    
            
        
        
        
    
        
                
            
        
        
        
    
            
            
        
    
    
        
    

        
        
    
                
                
                
                

        
            
            
                
            
                
    
 
        
    
    
    
    




    


   
        
    
    
    
    




    


