#include <bits/stdc++.h>
using namespace std;
int f(int k) {
  if (k < 4) return -1;
  if (k % 4 == 0) return (k / 4);
  if ((k - 6 >= 0) && ((k - 6) % 4 == 0)) return ((k - 6) / 4 + 1);
  if (((k - 9) >= 0) && ((k - 9) % 4 == 0)) return ((k - 9) / 4 + 1);
  if ((k - 15 >= 0) && ((k - 15) % 4 == 0)) return ((k - 15) / 4 + 2);
  return -1;
}
int main() {
  int n;
  int k;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> k;
    cout << f(k) << endl;
  }
}

### Prompt

Please formulate a cpp solution to the following problem:

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;
int f(int k) {
  if (k < 4) return -1;
  if (k % 4 == 0) return (k / 4);
  if ((k - 6 >= 0) && ((k - 6) % 4 == 0)) return ((k - 6) / 4 + 1);
  if (((k - 9) >= 0) && ((k - 9) % 4 == 0)) return ((k - 9) / 4 + 1);
  if ((k - 15 >= 0) && ((k - 15) % 4 == 0)) return ((k - 15) / 4 + 2);
  return -1;
}
int main() {
  int n;
  int k;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> k;
    cout << f(k) << endl;
  }
}

```