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;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
vector<long long int> v;
vector<long long int> w, ab, bc;
set<long long int> uo;
map<long long int, long long int> ww, cc, u;
vector<pair<long long int, long long int> > uu;
map<long long int, vector<long long int> > uuu;
map<pair<long long int, long long int>, long long int> vo;
map<pair<long long int, long long int>, long long int> vu;
long long int br[10005];
long long int dxx[] = {-1, 0, 1, 0};
long long int dyy[] = {0, 1, 0, -1};
long long int dx[] = {2, 1, -1, -2, -2, -1, 1, 2};
long long int dy[] = {1, 2, 2, 1, -1, -2, -2, -1};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  char ch;
  string a, b, c;
  double x, y, d, e, f, g, h;
  long long int n, m, i, j, k, p, q, o, l, s, tt = 1, t, z, aa, r = 0;
  for (i = 1, l = 0; i <= 100000000000000000; i *= 6, l++) {
    v.push_back(i);
  }
  cin >> tt;
  while (tt--) {
    cin >> n;
    for (i = 0, k = 4e18; i < v.size(); i++) {
      if (v[i] >= n) {
        s = v[i];
        t = 0;
        while (1) {
          if (s > n && s % 2 == 0) {
            t++;
            s /= 2;
          } else {
            break;
          }
        }
        if (s == n) {
          k = min(i + t, k);
        }
      }
    }
    if (k == 4e18) {
      cout << -1 << "\n";
    } else {
      cout << k << "\n";
    }
  }
  return 0;
}
