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;
const long long int INF = 2e18;
const long long int N = 2e5 + 5;
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6) {
    if (n % i == 0 || n % (i + 2) == 0) return false;
  }
  return true;
}
long long int exp(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n % 2 == 1) res = (res * x) % 1000000007;
    x = (x * x) % 1000000007;
    n = n / 2;
  }
  return res;
}
void null() {
  long long int n;
  cin >> n;
  vector<long long int> v;
  map<long long int, long long int> m;
  if (n == 1) {
    cout << 0 << '\n';
    return;
  }
  if (n == 3) {
    cout << 2 << '\n';
    return;
  }
  while (n % 2 == 0) {
    m[2]++;
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      if (i != 3) {
        cout << -1 << '\n';
        return;
      }
      n = n / i;
      m[3]++;
    }
  }
  if (m[2] > m[3]) {
    cout << -1 << '\n';
    return;
  }
  long long int z = m[3] - m[2];
  cout << m[3] + z << '\n';
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  clock_t start, end;
  start = clock();
  cout << fixed << setprecision(15);
  cin >> t;
  while (t--) {
    null();
  }
  end = clock();
  double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
}
