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

from sys import stdin, stdout
import math
cin = stdin.readline
cout = stdout.write
mp = lambda:list(map(int, cin().split()))

def mlog(a, b):
	x = 0
	while not a%b:
		x += 1
		a //= b
	return x
	'''
	x = int(math.log(a, b))
	if b**x == a:
		return x
	return x+1
	'''

t, = mp()
for _ in range(t):
	n, = mp()
	
	#y = n
	#z = n
	count2 = mlog(n, 2)
	count3 = mlog(n, 3)
	
	if count2 > count3:
		cout('-1\n')
	else:
		if 2**count2 * 3** count3 == n:
			cout(str(count3*2 - count2) + '\n')
		else:
			cout('-1\n')
	#print(count2, count3)
	#print(y,z)
	#if n*(2**z) == 6**y:
	#	cout(str(y+z) + '\n')
	#else:
	#	#print(y, z)
	#	cout('-1\n')
	