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 powoftwonumber(long long, long long);
long long sumofprimenumber(long long);
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, flag = 1, e = 1, count = 0;
    cin >> n;
    long long k = n;
    if (n == 1) {
      cout << "0"
           << "\n";
    } else {
      while (n != 1) {
        while (n % 6 == 0) {
          e = 0;
          n = n / 6;
          count++;
        }
        if (n % 6 == 3 && n < 10) {
          long long h = n / 6;
          count = count + 2 * (h + 1);
          flag = 1;
          break;
        }
        if (n % 6 == 2 || n % 6 == 4) {
          flag = 0;
          break;
        } else {
          n = n * 2;
          count++;
        }
        if (n > k * 6) {
          flag = 0;
          break;
        }
      }
      if (flag == 0) {
        cout << "-1"
             << "\n";
      } else {
        cout << count << "\n";
      }
    }
  }
  return 0;
}
long long sumofprimenumber(long long N) {
  long long i, Number, count, s = 0;
  for (Number = 2; Number <= N; Number++) {
    count = 0;
    for (i = 2; i <= sqrt(Number); i++) {
      if (Number % i == 0) {
        count++;
        break;
      }
    }
    if (count == 0) {
      s = s + Number;
    }
  }
  return s;
}
long long powoftwonumber(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if ((b & 1) != 0) {
      res = (res % 1000000007 * a % 1000000007) % 1000000007;
    }
    a = (a % 1000000007 * a % 1000000007) % 1000000007;
    b = b >> 1;
  }
  return res;
}
