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>
template <typename T>
constexpr int len(const T& container) {
  return container.size();
}
using ll = long long;
using namespace std;
int INF = 1e9 + 5;
pair<ll, ll> dirs[4] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
template <typename T, typename A>
int argmax(std::vector<T, A> const& vec) {
  return static_cast<int>(
      std::distance(vec.begin(), max_element(vec.begin(), vec.end())));
}
template <typename T, typename A>
int argmin(std::vector<T, A> const& vec) {
  return static_cast<int>(
      std::distance(vec.begin(), min_element(vec.begin(), vec.end())));
}
ll N = 0;
unordered_map<ll, ll> memo;
ll brute(ll n) {
  if (memo.find(n) != memo.end()) return memo[n];
  if (n == 1) return 0;
  if (n > 6 * N) return INF;
  ll result = INF;
  if (n % 6 == 0) result = min(result, 1 + brute(n / 6));
  result = min(result, 1 + brute(2 * n));
  memo[n] = result;
  return result;
}
int main() {
  ll t;
  cin >> t;
  for (auto i = 0; i < t; ++i) {
    ll n;
    cin >> n;
    N = n;
    auto result = brute(n);
    if (result >= INF) result = -1;
    cout << result << "\n";
  }
}
