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, i, n1, n, k, k1, j;
  cin >> t;
  vector<long long> ans, m;
  for (i = 1; i <= t; i++) {
    k = 0;
    k1 = 0;
    cin >> n;
    n1 = n;
    if (n == 1) {
      ans.push_back(0);
      continue;
    }
    while (n1 > 1) {
      for (j = 3; j >= 1; j--)
        if (n1 % j == 0) {
          if (j == 3) {
            k++;
            n1 = n1 / j;
            break;
          }
          if (j == 2) k1++;
          n1 = n1 / j;
          break;
        }
      if (k == 0 or j == 1) break;
    }
    if (k == 0 or k < k1 or j == 1) {
      ans.push_back(-1);
      continue;
    }
    if (k >= k1) ans.push_back(k + k - k1);
  }
  for (i = 0; i < ans.size(); i++) cout << ans[i] << endl;
}
