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

import sys
import math
# sys.stdin=open('input.txt','r')
# sys.stdout=open('output2.txt','w')
from sys import stdin,stdout
from collections import deque,defaultdict
from math import ceil,floor,inf,sqrt,factorial,gcd,log2
from copy import deepcopy
ii1=lambda:int(stdin.readline().strip())
is1=lambda:stdin.readline().strip()
iia=lambda:list(map(int,stdin.readline().strip().split()))
isa=lambda:stdin.readline().strip().split()
mod=1000000007
def factor(n):
    arr=[0]*2
    while n%2==0:
        arr[0]+=1
        n//=2
    while n%3==0:
        arr[1]+=1
        n//=3
    if n>1:
        arr.append(n)
    return arr
for _ in range(ii1()):
    n = ii1()
    if n==1:
        print(0)
        continue
    if n==3:
        print(2)
        continue
    if bin(n).count('1')==1:
        print(-1)
        continue
    arr=factor(n)
    if len(arr)==2 and (arr[0]>0 or arr[1]>0):
        # print(arr)
        if arr[0]==arr[1]:
            print(arr[0])
        elif arr[0]>arr[1]:
            print(-1)
        else:
            print(arr[1]+arr[1]-arr[0])
    else:
        print(-1)
