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

#!/usr/bin/python3

import getopt
import sys
import math

if __name__ == "__main__":
    onlyCase = -1
    verbose = False
    
    try:
         opts, args = getopt.getopt(sys.argv[1:], "vc:",
                                   ["case="])
    except getopt.GetoptError as err:
        print (str(err)) 
        sys.exit(2)

    for o, a in opts:
        if o in ("-c", "--case"):
            onlyCase = int(a)
        elif o  == "-v":
            verbose = True
        else: sys.exit()

    
    ncases = int(input())
    for c in range(1, ncases + 1):
        n = int(input())

        if onlyCase != -1 and onlyCase != c:
            continue

        # find how often divisible by 3 and 2
        d3 = 0
        d2 = 0

        x = n
        while x % 3 == 0:
            x = x // 3
            d3 += 1

        while x % 2 == 0:
            x = x // 2
            d2 += 1

        if verbose:
            print("%d by 3: %d by 2 %d (left: %d)" % (n,d3,d2, x)) 

        if d2 > d3 or x != 1:
            print("-1")
        else:
            print("%d" % (2*d3 - d2))





        

        




