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

t=int(input())

while t:
    n=int(input())
    
    a=0
    b=0
    while(n%2==0):
        a+=1
        n=n//2
 
    while (n%3==0):
        b+=1
        n=n//3
 
    if n==1:
        
        if a>b:
            print(-1)
 
        else:
            x = abs(b-a)
            print(max(b,a) + x)
    else:
        print(-1)
    
    t-=1
 

##for i in range(n):
##    currentTestcase = int(input())
##    counter = 0
##    appended = False
##    if currentTestcase == 1:
##        output.append(0)
##        appended = True
##    elif currentTestcase % 3 != 0:
##        output.append(-1)
##        appended = True
##    while appended == False:
##        counter += 1
##        if currentTestcase % 2 == 0 and currentTestcase % 3 != 0:
##            appended = True
##            output.append(-1)
##            break
##        elif currentTestcase % 6 == 0:
##            currentTestcase = currentTestcase//6
##            if currentTestcase == 1:
##                break
##            else:
##                continue
##        elif currentTestcase % 2 != 0 and currentTestcase % 3 == 0:
##            currentTestcase = currentTestcase * 2
##            continue
##        elif currentTestcase % 2 != 0 and currentTestcase % 3 != 0:
##            appended = True
##            output.append(-1)
##    if counter != 0 and appended == False:
##        output.append(counter)
##for i in output:
##    print(i)
