#include <bits/stdc++.h>
using namespace std;
int main(int argc, char** argv) {
  int n;
  cin >> n;
  while (n--) {
    int b, k;
    cin >> b;
    if (b % 4 == 0)
      k = b / 4;
    else if (b % 4 == 1) {
      if (b / 4 >= 2)
        k = b / 4 - 1;
      else
        k = -1;
    } else if (b % 4 == 2) {
      if (b / 4 >= 1)
        k = b / 4;
      else
        k = -1;
    } else if (b % 4 == 3) {
      if (b / 4 >= 3)
        k = b / 4 - 1;
      else
        k = -1;
    }
    cout << k << endl;
  }
  return 0;
}

### 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;
int main(int argc, char** argv) {
  int n;
  cin >> n;
  while (n--) {
    int b, k;
    cin >> b;
    if (b % 4 == 0)
      k = b / 4;
    else if (b % 4 == 1) {
      if (b / 4 >= 2)
        k = b / 4 - 1;
      else
        k = -1;
    } else if (b % 4 == 2) {
      if (b / 4 >= 1)
        k = b / 4;
      else
        k = -1;
    } else if (b % 4 == 3) {
      if (b / 4 >= 3)
        k = b / 4 - 1;
      else
        k = -1;
    }
    cout << k << endl;
  }
  return 0;
}

```