#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  int i, q, n, an;
  int a[16];
  for (i = 0; i < 16; i++) a[i] = -1;
  a[4] = 1;
  a[6] = 1;
  a[8] = 2;
  a[9] = 1;
  a[10] = 2;
  a[12] = 3;
  a[13] = 2;
  a[14] = 3;
  a[15] = 2;
  cin >> q;
  while (q--) {
    cin >> n;
    if (n < 16) {
      cout << a[n];
    } else {
      an = 0;
      an = n / 4;
      an -= 3;
      n %= 4;
      n += 12;
      cout << (an + a[n]);
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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() {
  ios::sync_with_stdio(false);
  int i, q, n, an;
  int a[16];
  for (i = 0; i < 16; i++) a[i] = -1;
  a[4] = 1;
  a[6] = 1;
  a[8] = 2;
  a[9] = 1;
  a[10] = 2;
  a[12] = 3;
  a[13] = 2;
  a[14] = 3;
  a[15] = 2;
  cin >> q;
  while (q--) {
    cin >> n;
    if (n < 16) {
      cout << a[n];
    } else {
      an = 0;
      an = n / 4;
      an -= 3;
      n %= 4;
      n += 12;
      cout << (an + a[n]);
    }
    cout << endl;
  }
  return 0;
}

```