#include <bits/stdc++.h>
using namespace std;
inline long long ri() {
  long long x;
  scanf("%lld", &x);
  return x;
}
template <typename T>
inline bool smax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
inline bool smin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < (n); i++) {
    int x;
    cin >> x;
    int res = -1;
    for (int i = 0; i < 4; i++)
      for (int j = 0; j < 4; j++) {
        int s = i * 6 + j * 9;
        if (s <= x && (x - s) % 4 == 0) {
          smax(res, (i + j + (x - s) / 4));
        }
      }
    cout << res << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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;
inline long long ri() {
  long long x;
  scanf("%lld", &x);
  return x;
}
template <typename T>
inline bool smax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
inline bool smin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < (n); i++) {
    int x;
    cin >> x;
    int res = -1;
    for (int i = 0; i < 4; i++)
      for (int j = 0; j < 4; j++) {
        int s = i * 6 + j * 9;
        if (s <= x && (x - s) % 4 == 0) {
          smax(res, (i + j + (x - s) / 4));
        }
      }
    cout << res << endl;
  }
  return 0;
}

```