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() {
  vector<unsigned long long int> six;
  unsigned long long int x = 6;
  while (x < 1e18) {
    six.push_back(x);
    x *= 6;
  }
  cout << six.size();
  int t;
  cin >> t;
  while (t--) {
    unsigned long long int n;
    cin >> n;
    int ans = 0, fl = 0;
    unsigned long long int val = 0;
    vector<unsigned long long int> two;
    unsigned long long int m = n;
    while (m < 1e18) {
      two.push_back(m);
      m *= 2;
    }
    int flag = 0;
    for (int i = 0; i < two.size(); i++) {
      if (binary_search(six.begin(), six.end(), two[i])) {
        flag = 1;
        val = two[i];
        break;
      }
    }
    if (!flag) {
      cout << -1 << "\n";
      continue;
    }
    for (int i = 0; i < two.size(); i++) {
      if (two[i] == val) {
        ans = i;
        break;
      }
    }
    for (int i = 0; i < six.size(); i++) {
      if (six[i] == val) {
        ans += i + 1;
        break;
      }
    }
    cout << ans << "\n";
  }
  return 0;
}
