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


# MasterKali

from sys import stdin
from collections import Counter
from math import sqrt, factorial, log10, log, floor, ceil
from itertools import permutations, combinations, combinations_with_replacement

input = stdin.readline
def li(): return list(map(int, input().split()))
def lis(): return list(map(str, input().split()))
def mp(): return map(int, input().split())
def inp(): return int(input())
def inps(): return str(input().strip())
def pr(n): return stdout.write(str(n)+"\n")

INF = float('inf')


def solve():
	n = inp()
	if n == 1: return 0
	if n == 2 or n%5==0: return -1
	k, cnt = 0, 0
	while (n!=1):
		if n%6==0:
			n//=6
		else:
			x = n%6
			if x==2: cnt+=1
			if x == 2 and cnt == 2:
				return -1
			n*=2
		if n == 2:
			return -1
		k+=1		

	return k


t = inp()
for i in range(t):
	print(solve())

