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;
void countSort(vector<int>& a, int k) {
  vector<int> count(k + 1, 0);
  int n = a.size();
  for (int i : a) count[i]++;
  int num_so_far = 0;
  for (int i = 0; i <= k; i++) {
    int c = count[i];
    count[i] = num_so_far;
    num_so_far += c;
  }
  vector<int> output(n, 0);
  for (int i : a) {
    output[count[i]] = i;
    count[i]++;
  }
  for (int i = 0; i < n; i++) a[i] = output[i];
}
void countSort2(vector<int>& a) {
  int mn = *min_element(a.begin(), a.end());
  int mx = *max_element(a.begin(), a.end());
  int range = mx - mn + 1;
  for (int i = 0; i < a.size(); i++) a[i] -= mn;
  countSort(a, range);
  for (int i = 0; i < a.size(); i++) a[i] += mn;
}
long solve(long n) {
  long two = 0;
  long three = 0;
  long t = n;
  while (t % 2 == 0) {
    two++;
    t = t / 2;
  }
  while (t % 3 == 0) {
    three++;
    t = t / 3;
  }
  if (t != 1 or two > three or three == 0) {
    return -1;
  }
  if (two == three) return two;
  return three - two + three;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long n;
    cin >> n;
    cout << solve(n) << endl;
  }
}
