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


# Problem B: Multiply by 2, divide by 6

import sys

caseCount = int(next(sys.stdin))

def v(p, n):
    lo = 0
    hi = 30
    
    while hi > lo + 1:
        mid = (lo + hi) // 2
        if n % (p ** mid) == 0:
            lo = mid
        else:
            hi = mid - 1
            
    if n % (p ** hi) == 0:
        return hi
    return lo

def solve(n):
    threes = v(3, n)
    twos = v(2, n)
    
    if threes < twos:
        return -1
        
    if (3 ** threes) * (2 ** twos) != n:
        return -1
        
    return threes + (threes - twos)


ans = ""

for i in range(caseCount):
    n = int(next(sys.stdin))
    ans += str(solve(n)) + "\n"
    
print(ans)