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

#pragma GCC target("avx2")
#pragma GCC optimize("O3")


#include<bits/stdc++.h>
#include <ext/pb_ds/detail/standard_policies.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>


#define ll long long
#define f first
#define die exit(0)
#define s second
#define pb push_back
#define pf push_front
#define pp pop_back
#define ld long double

using namespace std;
using namespace __gnu_pbds;

typedef tree<ll , null_type,
less<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
/*
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*/
mt19937 gen(time(0));
int32_t main()
{

    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#else
    // freopen("cubes.in", "r", stdin);
    // freopen("cubes.out", "w", stdout);
#endif // LOCAL



  ll t;
  cin>> t;
  while(t--) {
    ll n;
    cin >> n;
    string s;
    cin >> s;
    ll nom = 0;
    ll mi = 0;
    for(int i = 0; i < s.size(); i++) {
        if(s[i] == '(') nom++;
        else nom--;
        mi = min(mi, nom);
    }
    cout << abs(mi) << '\n';
  }
}
