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;
bool trash(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
void solve() {
  long long int a, b, c, n, nn, m, t, k, r, x, y, z, o, l,
      dd = 1000000, rr, i, sl, tl, d, mid, val, j, lo, h, e, tmp, cn, cf, flag;
  long long int ans = 0, sum = 0, two = 0, three = 0;
  cin >> x;
  if (x == 1) {
    cout << 0 << "\n";
  } else if (x == 3) {
    cout << 2 << "\n";
  } else if (x < 6) {
    cout << -1 << "\n";
  } else {
    while (x % 2 == 0) {
      x = x / 2;
      two++;
    }
    while (x % 3 == 0) {
      x /= 3;
      three++;
    }
    if (x != 1 or three < two)
      cout << -1 << "\n";
    else
      cout << (2 * three) - two << "\n";
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
