#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long sum = 0, flag = 0, i, j, k, x, y, z, n, m, p, t;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> x;
    if (x % 4 == 0)
      cout << x / 4;
    else if ((x - 6) % 4 == 0 && (x - 6) >= 0) {
      cout << (x - 6) / 4 + 1;
    } else if ((x - 9) % 4 == 0 && (x - 9) >= 0) {
      cout << (x - 9) / 4 + 1;
    } else if ((x - 15) % 4 == 0 && (x - 15) >= 0) {
      cout << (x - 15) / 4 + 2;
    } else
      cout << -1;
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t;
  t = 1;
  for (int r = 0; r < t; r++) {
    solve();
  }
  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;
void solve() {
  long long sum = 0, flag = 0, i, j, k, x, y, z, n, m, p, t;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> x;
    if (x % 4 == 0)
      cout << x / 4;
    else if ((x - 6) % 4 == 0 && (x - 6) >= 0) {
      cout << (x - 6) / 4 + 1;
    } else if ((x - 9) % 4 == 0 && (x - 9) >= 0) {
      cout << (x - 9) / 4 + 1;
    } else if ((x - 15) % 4 == 0 && (x - 15) >= 0) {
      cout << (x - 15) / 4 + 2;
    } else
      cout << -1;
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t;
  t = 1;
  for (int r = 0; r < t; r++) {
    solve();
  }
  return 0;
}

```