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

###DO NOT KNOW THE ANSWER...I AM REALLY SAD###

query=int(input())
for _ in range(query):
	n=int(input())
	c3=c2=0	#Counts how many times the number can be divided by 3 and 2
	#n=2**c2*3**c3
	#if needs to be made c3=0; n=2**(c2-c3)*3**0
	while(n%3==0):
		n//=3
		c3+=1
	while(n%2==0):
		n//=2
		c2+=1
	if c2>c3:	#if c2>c3 then there will be no 6 left to divide, only 2 can be mutiplied, if c2>c3 division by 2 is implied
		print(-1)
	elif n!=1:	#if not n==1
		print(-1)
	else:
		print(c3+(c3-c2))	#c3-c2 means how many times 2 is to be multiplied

# Video ref: https://www.youtube.com/watch?v=ekxYAYio9Xg
#Video 2: https://www.youtube.com/watch?v=sPl79SlXwRc