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>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using lld = long double;
const ll oo = 0x3f3f3f3f;
const ll MOD = 1000000007;
const ll N = 1e9 + 10;
map<ll, ll> dis;
inline void insert(const ll& x, const ll& p, queue<ll>& q) {
  if (x >= N) return;
  if (!dis.count(x)) {
    q.push(x);
    dis[x] = 1 + dis[p];
  }
}
void bfs(ll src = 1) {
  queue<ll> q;
  q.push(src);
  dis[src] = 0;
  while (!q.empty()) {
    ll t = q.front();
    q.pop();
    if (t % 2 == 0) insert(t / 2, t, q);
    insert(t * 6, t, q);
  }
}
void solve() {
  ll n;
  cin >> n;
  if (!dis.count(n))
    cout << "-1\n";
  else
    cout << dis[n] << '\n';
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  bfs();
  ll tc;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}
