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 mex(long long int a, long long int b, long long int m) {
  if (b == 0) return 1;
  if (b % 2)
    return a * mex((a * a) % m, (b - 1) / 2, m) % m;
  else
    return mex((a * a) % m, b / 2, m);
}
int x, y;
void exteu(long long int a, long long int m) {
  if (m == 0) {
    x = 1;
    y = 0;
  } else {
    exteu(m, a % m);
    long long int temp = x;
    x = y;
    y = temp - (a / m) * y;
  }
}
long long int fact(long long int n) {
  long long int prod = 1;
  for (long long int i = 2; i <= n; i++) {
    prod *= i;
  }
  return prod;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void solve() {
  long long int n;
  cin >> n;
  if (n == 1) {
    cout << 0 << endl;
    return;
  }
  long long int p3 = 0, p2 = 0;
  for (int i = 2; i <= sqrt(n); ++i) {
    while (n % i == 0 && n) {
      n /= i;
      if (i == 2)
        p2++;
      else if (i == 3)
        p3++;
      else {
        cout << -1 << endl;
        return;
      }
    }
  }
  if (n != 1) {
    if (n == 2) p2++;
    if (n == 3) p3++;
  }
  if (p3 != 0 && p3 >= p2) {
    cout << (p3 - p2) + p3 << endl;
  } else
    cout << -1 << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int TESTS = 1;
  cin >> TESTS;
  while (TESTS--) {
    solve();
  }
  return 0;
}
