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 modp(long long int a, long long int p, long long int n) {
  long long int d[100];
  long long int i, j, k, l;
  if (n == 0) return 1;
  for (i = 0; n >= 1; i++) {
    d[i] = n % 2;
    n /= 2;
  }
  l = i;
  long long int e = 1;
  if (d[0] == 1) e *= (a % p);
  for (i = 1; i < l; i++) {
    a *= a;
    a = a % p;
    if (d[i] == 1) {
      e *= a;
      e = e % p;
    }
  }
  return e % p;
}
long long int modInverse(long long int n, long long int p) {
  return modp(n, p, p - 2);
}
bool sorter(pair<long long int, long long int>& a,
            pair<long long int, long long int>& b) {
  if (a.first != b.first) {
    return a.first > b.first;
  }
  return b.second > a.second;
}
long long int p1 = 1e9 + 7, p2 = 998244353;
long long int l, r, mid, ans;
long long int n, i, j, k, g, m;
long long int x, y, n1, n2, h, z, c;
long long int aa = 1;
void solve() {
  cin >> n;
  x = 0;
  y = 0;
  while (n % 2 == 0) {
    x += 1;
    n /= 2;
  }
  while (n % 3 == 0) {
    y += 1;
    n /= 3;
  }
  if (n != 1 || x > y) {
    cout << "-1";
    return;
  }
  ans = 2 * y - x;
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  cin >> t;
  long long int ans = 0;
  for (int i = 0; i < t; i++) {
    solve();
    cout << "\n";
  }
  return 0;
}
