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;
const double pi = acos(-1.0);
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
int n;
long long int solve() {
  long long int res = 0;
  int pow2 = 0;
  int pow3 = 0;
  int n2 = n;
  int n2a = 1;
  while (n2 % 2 == 0) {
    n2 /= 2;
    pow2++;
    n2a *= 2;
  }
  int n3b = 1;
  int n3 = n;
  while (n3 % 3 == 0) {
    n3 /= 3;
    pow3++;
    n3b *= 3;
  }
  if (n2a * n3b != n) {
    return -1;
  }
  if (pow2 > pow3) {
    return -1;
  }
  res = pow3;
  res += pow3 - pow2;
  return res;
}
int main() {
  int T;
  cin >> T;
  for (int t = 0; t < T; t += 1) {
    cin >> n;
    cout << solve() << endl;
  }
  return 0;
}
