#include <bits/stdc++.h>
using namespace std;
int main() {
  int q;
  scanf("%d", &q);
  while (q--) {
    int n;
    scanf("%d", &n);
    bool flag = 0;
    if (n <= 3 || n == 5 || n == 7 || n == 11)
      ;
    else
      flag = 1;
    int ans;
    if (!flag)
      ans = -1;
    else {
      if (n % 4 == 0)
        ans = n / 4;
      else if (n % 4 == 1)
        ans = (n - 9) / 4 + 1;
      else if (n % 4 == 2)
        ans = (n - 6) / 4 + 1;
      else
        ans = (n - 6 - 9) / 4 + 2;
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Generate 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 main() {
  int q;
  scanf("%d", &q);
  while (q--) {
    int n;
    scanf("%d", &n);
    bool flag = 0;
    if (n <= 3 || n == 5 || n == 7 || n == 11)
      ;
    else
      flag = 1;
    int ans;
    if (!flag)
      ans = -1;
    else {
      if (n % 4 == 0)
        ans = n / 4;
      else if (n % 4 == 1)
        ans = (n - 9) / 4 + 1;
      else if (n % 4 == 2)
        ans = (n - 6) / 4 + 1;
      else
        ans = (n - 6 - 9) / 4 + 2;
    }
    printf("%d\n", ans);
  }
  return 0;
}

```