#include <bits/stdc++.h>
using namespace std;
int main() {
  int q;
  cin >> q;
  long long n[q];
  for (int i = 0; i < q; i++) {
    cin >> n[i];
  }
  long long ans[q];
  for (int i = 0; i < q; i++) {
    int div = n[i] % 4;
    if (div) {
      if (div == 1) ans[i] = (n[i] - 9) / 4 + 1;
      if (div == 2) ans[i] = (n[i] - 6) / 4 + 1;
      if (div == 3) ans[i] = (n[i] - 15) / 4 + 2;
      if (ans[i] < 1) ans[i] = -1;
      if (n[i] == 11) ans[i] = -1;
    } else
      ans[i] = n[i] / 4;
  }
  for (int i = 0; i < q; i++) {
    cout << ans[i] << "\n";
  }
}

### Prompt

Develop a solution in CPP 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 q;
  cin >> q;
  long long n[q];
  for (int i = 0; i < q; i++) {
    cin >> n[i];
  }
  long long ans[q];
  for (int i = 0; i < q; i++) {
    int div = n[i] % 4;
    if (div) {
      if (div == 1) ans[i] = (n[i] - 9) / 4 + 1;
      if (div == 2) ans[i] = (n[i] - 6) / 4 + 1;
      if (div == 3) ans[i] = (n[i] - 15) / 4 + 2;
      if (ans[i] < 1) ans[i] = -1;
      if (n[i] == 11) ans[i] = -1;
    } else
      ans[i] = n[i] / 4;
  }
  for (int i = 0; i < q; i++) {
    cout << ans[i] << "\n";
  }
}

```