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
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import ceil, floor, gcd, sqrt, trunc, inf
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var, end="\n"): sys.stdout.write(str(var)+end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]


"""
                            pppppppppppppppppppp
                         ppppp  ppppppppppppppppppp
                      ppppppp    ppppppppppppppppppppp
                      pppppppp  pppppppppppppppppppppp
                      pppppppppppppppppppppppppppppppp
                              pppppppppppppppppppppppp
       ppppppppppppppppppppppppppppppppppppppppppppppp  pppppppppppppppppppp
      pppppppppppppppppppppppppppppppppppppppppppppppp  ppppppppppppppppppppp
     ppppppppppppppppppppppppppppppppppppppppppppppppp  pppppppppppppppppppppp
    ppppppppppppppppppppppppppppppppppppppppppppppp    pppppppppppppppppppppppp
   pppppppppppppppppppppppppppppppppppppppppppppp     pppppppppppppppppppppppppp
  ppppppppppppppppppppppppppppppppppppppppppppp      pppppppppppppppppppppppppppp
  pppppppppppppppppppppppppppppppp               pppppppppppppppppppppppppppppppp
  pppppppppppppppppppppppppppp     pppppppppppppppppppppppppppppppppppppppppppppp
  ppppppppppppppppppppppppppp    pppppppppppppppppppppppppppppppppppppppppppppppp
    pppppppppppppppppppppppp  pppppppppppppppppppppppppppppppppppppppppppppppppp
     ppppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppppppp
      pppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppppp
       ppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppp
                              pppppppppppppppppppppppp
                              pppppppppppppppppppppppppppppppp
                              pppppppppppppppppppppp  pppppppp
                              ppppppppppppppppppppp    ppppppp
                                 ppppppppppppppppppp  ppppp
                                    pppppppppppppppppppp                                                         
"""

for _ in range(int(data())):
    n = int(data())
    if n == 1:
        out(0)
        continue
    if n % 3 != 0:
        out(-1)
        continue
    two, three = 0, 0
    while n % 2 == 0:
        two += 1
        n //= 2
    while n % 3 == 0:
        three += 1
        n //= 3
    if n > 1 or two > three:
        out(-1)
        continue
    out(2 * three - two)
