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>
#pragma GCC target("avx")
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
template <class T>
inline void in(T &);
template <class T>
inline void out(const T &);
using std::abs;
using std::max;
using std::min;
using std::sort;
using namespace std;
int t, n, s1, s2;
int main() {
  in(t);
  while (t--) {
    in(n);
    s1 = s2 = 0;
    while (n % 2 == 0) n /= 2, s1++;
    while (n % 3 == 0) n /= 3, s2++;
    if (n == 1 && s1 <= s2)
      out(2 * s2 - s1);
    else
      out(-1);
    putchar('\n');
  }
  return 0;
}
template <class T>
void in(T &Re) {
  T k = 0;
  char ch = getchar();
  int flag = 1;
  while (!(ch >= '0' && ch <= '9')) {
    if (ch == '-') {
      flag = -1;
    }
    ch = getchar();
  }
  while ((ch >= '0' && ch <= '9')) {
    k = (k << 1) + (k << 3) + ch - '0';
    ch = getchar();
  }
  Re = flag * k;
}
template <class T>
void out(const T &Wr) {
  if (Wr < 0) {
    putchar('-');
    out(-Wr);
  } else {
    if (Wr < 10) {
      putchar(Wr + '0');
    } else {
      out(Wr / 10);
      putchar((Wr % 10) + '0');
    }
  }
}
