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 cmp(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;
  else {
    return a.second < b.second;
  }
}
void printV(vector<int> nums) {
  for (int i : nums) {
    printf("%d ", i);
  }
  printf("\n");
}
int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
bool valid(int i, int j, int n, int m) {
  if (i >= 0 && i < n && j >= 0 && j < m)
    return true;
  else
    return false;
}
bool isPrime(int n) {
  if (n == 1) return false;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long int kadane(vector<int> nums) {
  int n = nums.size();
  long long int mx = 0;
  long long int curr = 0;
  for (int i = 0; i < n; i++) {
    curr += nums[i];
    if (curr < 0) curr = 0;
    mx = max(mx, curr);
  }
  return mx;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int cnt = 0;
    bool flag = true;
    while (n != 1) {
      if (n % 6 == 0)
        n /= 6;
      else if ((n * 2) % 6 == 0) {
        n *= 2;
      } else {
        flag = false;
        break;
      }
      cnt++;
    }
    if (!flag) {
      cout << -1 << endl;
    } else {
      cout << cnt << endl;
    }
  }
  return 0;
}
