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

#include <bits/stdc++.h>
template <typename T>
void scan(T &x) {
  x = 0;
  bool _ = 0;
  T c = getchar();
  _ = c == 45;
  c = _ ? getchar() : c;
  while (c < 48 || c > 57) c = getchar();
  for (; c < 48 || c > 57; c = getchar())
    ;
  for (; c > 47 && c < 58; c = getchar()) x = (x << 3) + (x << 1) + (c & 15);
  x = _ ? -x : x;
}
template <typename T>
void printn(T n) {
  bool _ = 0;
  _ = n < 0;
  n = _ ? -n : n;
  char snum[65];
  int i = 0;
  do {
    snum[i++] = n % 10 + 48;
    n /= 10;
  } while (n);
  --i;
  if (_) putchar(45);
  while (i >= 0) putchar(snum[i--]);
}
template <typename First, typename... Ints>
void scan(First &arg, Ints &...rest) {
  scan(arg);
  scan(rest...);
}
template <typename T>
void print(T n) {
  printn(n);
  putchar(10);
}
template <typename First, typename... Ints>
void print(First arg, Ints... rest) {
  printn(arg);
  putchar(32);
  print(rest...);
}
using namespace std;
int main() {
  int t, n;
  scan(t);
  while (t--) {
    scan(n);
    int cnt2 = 0, cnt3 = 0;
    while (n % 2 == 0) {
      cnt2++;
      n /= 2;
    }
    while (n % 3 == 0) {
      n /= 3;
      cnt3++;
    }
    if (n > 1 || cnt2 > cnt3) {
      print(-1);
      continue;
    }
    print(cnt3 + cnt3 - cnt2);
  }
}
