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;
int main() {
  long long t;
  cin >> t;
  for (long long p = 1; p <= t; p++) {
    long long angka;
    cin >> angka;
    if (angka == 1) {
      cout << "0\n";
      continue;
    }
    long long langkah = 0;
    long long div3 = 0;
    long long div2 = 0;
    while (angka % 6 == 0) {
      angka = angka / 6;
      langkah++;
    }
    while (angka % 3 == 0) {
      angka = angka / 3;
      div3++;
    }
    while (angka % 2 == 0) {
      angka = angka / 2;
      div2++;
    }
    if (angka == 1) {
      if (div3 == 0 and div2 == 0) {
        cout << langkah << "\n";
        continue;
      }
    }
    if (angka > 1) {
      cout << "-1\n";
    } else {
      if (div3 == 0) {
        cout << "-1\n";
      } else {
        if (div2 > div3) {
          cout << "-1\n";
        } else {
          long long surplus = div3 - div2;
          langkah = langkah + surplus + div3;
          cout << langkah << "\n";
        }
      }
    }
  }
  return 0;
}
