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 mod(long long x) {
  x = x % 1000000007;
  if (x < 0) x += 1000000007;
  return x;
}
int gcd(int a, int b) {
  if (b > a) swap(a, b);
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e == 1) return mod(b);
  long long res = power(b, e / 2);
  if (e % 2 == 0) return mod(res * res);
  return mod(res * mod(res * b));
}
bool compare(vector<int>& a, vector<int>& b) {
  if (a[1] == b[1]) return a[0] > b[0];
  return a[1] < b[1];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    if (n == 1) {
      cout << "0\n";
    } else {
      int two = 0, three = 0;
      int f = 1;
      while (n > 1) {
        if (n % 2 == 0) {
          while (n % 2 == 0) {
            two++;
            n = n / 2;
          }
        } else if (n % 3 == 0) {
          while (n % 3 == 0) {
            three++;
            n = n / 3;
          }
        } else {
          f = 0;
          break;
        }
      }
      if (f == 0) {
        cout << "-1\n";
      } else {
        if (two > three) {
          cout << "-1\n";
        } else {
          long long ans = two + (three - two) * 2;
          cout << ans << "\n";
        }
      }
    }
  }
  return 0;
}
