#include <bits/stdc++.h>
using namespace std;
template <typename tp>
inline void read(tp &x) {
  x = 0;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; x = x * 10 + c - '0', c = getchar())
    ;
}
int n, T;
int main() {
  cin >> T;
  for (; T--;) {
    read(n);
    int ans = 0;
    if (n == 1 || n == 2 || n == 3 || n == 5 || n == 7 || n == 11)
      printf("-1\n");
    else {
      int res = n;
      if (res % 4 == 1) {
        res = res - 9;
        ans++;
      }
      if (res % 4 == 2) {
        res -= 6;
        ans++;
      }
      if (res % 4 == 3) {
        res -= 15;
        ans += 2;
      }
      if (res % 4 == 0) {
        ans += res / 4;
      }
      cout << ans << "\n";
    }
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
template <typename tp>
inline void read(tp &x) {
  x = 0;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; x = x * 10 + c - '0', c = getchar())
    ;
}
int n, T;
int main() {
  cin >> T;
  for (; T--;) {
    read(n);
    int ans = 0;
    if (n == 1 || n == 2 || n == 3 || n == 5 || n == 7 || n == 11)
      printf("-1\n");
    else {
      int res = n;
      if (res % 4 == 1) {
        res = res - 9;
        ans++;
      }
      if (res % 4 == 2) {
        res -= 6;
        ans++;
      }
      if (res % 4 == 3) {
        res -= 15;
        ans += 2;
      }
      if (res % 4 == 0) {
        ans += res / 4;
      }
      cout << ans << "\n";
    }
  }
}

```