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() {
  int n = 0, count = 0;
  cin >> n;
  int* ptrvalue = new int[n];
  float* num = new float[n];
  float num1 = 0.0;
  for (int i = 0; i < n; i++) {
    cin >> ptrvalue[i];
    num[i] = (float)ptrvalue[i];
  }
  for (int i = 0; i < n; i++) {
    while (true) {
      if (ptrvalue[i] != 1 && ptrvalue[i] > 0) {
        if (ptrvalue[i] % 2) {
          ptrvalue[i] *= 2;
          count++;
        } else {
          num1 = (float)ptrvalue[i] / 6;
          ptrvalue[i] /= 6;
          count++;
          if (ptrvalue[i] == 0 || ptrvalue[i] < 0) {
            count = -1;
            cout << count << endl;
            count = 0;
            break;
          }
        }
      } else if (num[i] == 1.00) {
        cout << count << endl;
        count = 0;
        break;
      } else if (num1 != 1.00) {
        count = -1;
        cout << count << endl;
        count = 0;
        break;
      } else {
        cout << count << endl;
        count = 0;
        break;
      }
    }
  }
  delete ptrvalue;
  return 0;
}
