#include <bits/stdc++.h>
using namespace std;
const long long int N = 31;
const long long int MOD = 998244353;
long long int mod(long long int n) { return (n % MOD + MOD) % MOD; }
void solve() {
  long long int x;
  cin >> x;
  if (x % 4 == 0)
    cout << x / 4 << endl;
  else if (x % 4 == 1)
    (x - 9) / 4 < 0 ? cout << -1 << endl : cout << 1 + (x - 9) / 4 << endl;
  else if (x % 4 == 2)
    (x - 6) / 4 < 0 ? cout << -1 << endl : cout << 1 + (x - 6) / 4 << endl;
  else if (x % 4 == 3)
    (x - 15) / 4 < 0 ? cout << -1 << endl : cout << 2 + (x - 15) / 4 << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int T = 1;
  cin >> T;
  for (long long int i = 0; i < T; i++) solve();
}

### Prompt

Please provide a cpp coded solution to the problem described below:

You are given several queries. In the i-th query you are given a single positive integer ni. You are to represent ni as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.

An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself.

Input

The first line contains single integer q (1 ≤ q ≤ 105) — the number of queries.

q lines follow. The (i + 1)-th line contains single integer ni (1 ≤ ni ≤ 109) — the i-th query.

Output

For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.

Examples

Input

1
12


Output

3


Input

2
6
8


Output

1
2


Input

3
1
2
3


Output

-1
-1
-1

Note

12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.

8 = 4 + 4, 6 can't be split into several composite summands.

1, 2, 3 are less than any composite number, so they do not have valid splittings.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long int N = 31;
const long long int MOD = 998244353;
long long int mod(long long int n) { return (n % MOD + MOD) % MOD; }
void solve() {
  long long int x;
  cin >> x;
  if (x % 4 == 0)
    cout << x / 4 << endl;
  else if (x % 4 == 1)
    (x - 9) / 4 < 0 ? cout << -1 << endl : cout << 1 + (x - 9) / 4 << endl;
  else if (x % 4 == 2)
    (x - 6) / 4 < 0 ? cout << -1 << endl : cout << 1 + (x - 6) / 4 << endl;
  else if (x % 4 == 3)
    (x - 15) / 4 < 0 ? cout << -1 << endl : cout << 2 + (x - 15) / 4 << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int T = 1;
  cin >> T;
  for (long long int i = 0; i < T; i++) solve();
}

```