#include <bits/stdc++.h>
using namespace std;
int n, r;
int p[10001], num[10000], t;
int a[100] = {-1, -1, -1, -1, 1, -1, 1, -1, 2, 1, 2, -1, 3};
int main() {
  for (int i = 2; i <= 10000; i++)
    if (!p[i]) {
      for (int j = i * 2; j <= 10000; j += i) p[j] = 1;
      num[t++] = i;
    }
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d", &r);
    if (r >= 13) {
      int ans = 0;
      if (r % 2 != 0) {
        r -= 9;
        ans++;
      }
      if (r % 4 != 0) {
        r -= 6;
        ans++;
      }
      ans += r / 4;
      cout << ans << endl;
    } else
      cout << a[r] << endl;
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 n, r;
int p[10001], num[10000], t;
int a[100] = {-1, -1, -1, -1, 1, -1, 1, -1, 2, 1, 2, -1, 3};
int main() {
  for (int i = 2; i <= 10000; i++)
    if (!p[i]) {
      for (int j = i * 2; j <= 10000; j += i) p[j] = 1;
      num[t++] = i;
    }
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d", &r);
    if (r >= 13) {
      int ans = 0;
      if (r % 2 != 0) {
        r -= 9;
        ans++;
      }
      if (r % 4 != 0) {
        r -= 6;
        ans++;
      }
      ans += r / 4;
      cout << ans << endl;
    } else
      cout << a[r] << endl;
  }
  return 0;
}

```