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>
using namespace std;
long long mpow(long long base, long long exp) {
  base %= 1000000007;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((long long)result * base) % 1000000007;
    base = ((long long)base * base) % 1000000007;
    exp >>= 1;
  }
  return result;
}
long long factorial(long long n) {
  long long i = n, fact = 1;
  while (n / i != n) {
    fact = fact * i;
    i--;
  }
  return fact;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long n, c = 0;
    cin >> n;
    if (n < 6) {
      if (n == 1)
        cout << 0 << endl;
      else if (n == 3)
        cout << 2 << endl;
      else
        cout << -1 << endl;
    } else if (n == 6)
      cout << 1 << endl;
    else {
      if (n % 6 == 0 && (n / 6) % 3 == 0) {
        while (n > 1) {
          n = n / 6;
          c++;
          if (n % 6 != 0) {
            n = n * 2;
            c++;
          }
        }
        cout << c - 2 << endl;
      } else {
        n = n * 2;
        c++;
        if (n % 6 == 0 && (n / 6) % 3 == 0) {
          while (n > 1) {
            n = n / 6;
            c++;
            if (n % 6 != 0) {
              n = n * 2;
              c++;
            }
          }
          cout << c - 2 << endl;
        } else
          cout << -1 << endl;
      }
    }
  }
  return 0;
}
