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;
using ll = long long;
using ld = long double;
using pr = pair<int, int>;
using prll = pair<long, long>;
using V = vector<int>;
using Vll = vector<long long>;
template <typename T, typename TT>
pr operator+(pair<T, TT> a, pair<T, TT> b) {
  return {a.first + b.first, a.second + b.second};
}
template <typename T, typename TT>
ostream& operator<<(ostream& s, pair<T, TT> t) {
  return s << "(" << t.first << "," << t.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, vector<T> t) {
  for (auto itr : t) s << itr << " ";
  return s;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    int x = 0;
    while (n > 1 && n % 6 == 0) {
      n /= 6;
      x++;
    }
    int ans = x;
    int y = 0;
    while (n > 1 && n % 3 == 0) {
      n /= 3;
      y++;
    }
    ans += 2 * y;
    if (n == 1)
      cout << ans << "\n";
    else
      cout << "-1\n";
  }
  return 0;
}
