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;
vector<long long int> v(500001, 0);
struct node {
  int a, b;
};
long long int min(long long int a, long long int b) { return a < b ? a : b; }
long long int max(long long int a, long long int b) { return a > b ? a : b; }
map<int, int> M;
void factorize(long long n) {
  int count = 0;
  while (!(n % 2)) {
    n >>= 1;
    count++;
  }
  if (count) M[2] = count;
  for (long long i = 3; i <= sqrt(n); i += 2) {
    count = 0;
    while (n % i == 0) {
      count++;
      n = n / i;
    }
    if (count) {
      M[i] = count;
    }
  }
  if (n > 2) {
    M[n] = 1;
  }
}
void test(int r) {
  long long int n;
  cin >> n;
  if (n == 1) {
    cout << -1 << endl;
  } else {
    M.erase(M.begin(), M.end());
    factorize(n);
    bool flag = true;
    long long int two = 0, three = 0;
    for (auto itr = M.begin(); itr != M.end(); itr++) {
      if (itr->first == 2) {
        two = itr->second;
      } else if (itr->first == 3) {
        three = itr->second;
      } else {
        flag = false;
        break;
      }
    }
    if (flag) {
      int small = min(two, three);
      if (small == two) {
        long long int a = three - two;
        long long int count = three + a;
        cout << count << endl;
      } else {
        cout << -1 << endl;
      }
    } else {
      cout << -1 << endl;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1, r;
  cin >> t;
  for (r = 1; r <= t; r++) test(r);
  return 0;
}
