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;
void allprime(int n, bool prime[]) {
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
void bfs(vector<int> adj[], int src, bool vis[], int level[]) {
  queue<pair<int, int> > q;
  q.push(make_pair(src, 0));
  vis[src] = true;
  level[src] = 0;
  while (!q.empty()) {
    pair<int, int> vpair = q.front();
    q.pop();
    int v = vpair.first;
    int lvl = vpair.second;
    for (int i = 0; i < adj[v].size(); i++) {
      int next = adj[v][i];
      if (!vis[next]) {
        q.push(make_pair(next, lvl + 1));
        vis[next] = true;
        level[next] = lvl + 1;
      }
    }
  }
}
void dfs(vector<int> adj[], int src, bool vis[]) {
  vis[src] = true;
  for (int i = 0; i < adj[src].size(); i++) {
    int next = adj[src][i];
    if (!vis[next]) {
      dfs(adj, next, vis);
    }
  }
}
void solve() {
  int n;
  cin >> n;
  int pow2 = 0, pow3 = 0;
  while (n % 2 == 0) {
    n /= 2;
    pow2++;
  }
  while (n % 3 == 0) {
    n /= 3;
    pow3++;
  }
  if (n != 1)
    cout << -1 << endl;
  else if (pow3 < pow2)
    cout << -1 << endl;
  else
    cout << 2 * pow3 - pow2 << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}
