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 int power(long long int x, long long int y, long long int m) {
  if (y == 0) return 1;
  long long int p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long int nCr(long long int n, long long int r, long long int m) {
  if (r > n) return 0;
  long long int a = 1, b = 1, i;
  for (i = 0; i < r; i++) {
    a = (a * n) % m;
    --n;
  }
  while (r) {
    b = (b * r) % m;
    --r;
  }
  return (a * power(b, m - 2, m)) % m;
}
void solve() {
  long long int n;
  cin >> n;
  int t = 0, s = 0;
  while (n % 2 == 0) {
    n /= 2;
    t++;
  }
  while (n % 3 == 0) {
    n /= 3;
    s++;
  }
  if (n != 1) {
    cout << -1 << '\n';
    return;
  }
  if (t == s)
    cout << t << '\n';
  else if (t < s)
    cout << 2 * s - t << '\n';
  else
    cout << -1 << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
