#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
const double eps = 1e-7;
const double pi = 3.14159265358979323846264338327950288L;
template <class T>
inline void chkmax(T &x, T y) {
  if (x < y) x = y;
}
template <class T>
inline void chkmin(T &x, T y) {
  if (x > y) x = y;
}
int t, n;
int calc(int n) {
  if (n % 2 == 0 && n > 2) return n / 4;
  if (n == 9) return 1;
  if (n % 2 == 1 && n > 9) return calc(n - 9) + 1;
  return -inf;
}
int main() {
  scanf("%d", &t);
  while (t--) scanf("%d", &n), printf("%d\n", max(calc(n), -1));
  return 0;
}

### Prompt

Please create a solution in cpp 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;
const int inf = 1e9 + 7;
const double eps = 1e-7;
const double pi = 3.14159265358979323846264338327950288L;
template <class T>
inline void chkmax(T &x, T y) {
  if (x < y) x = y;
}
template <class T>
inline void chkmin(T &x, T y) {
  if (x > y) x = y;
}
int t, n;
int calc(int n) {
  if (n % 2 == 0 && n > 2) return n / 4;
  if (n == 9) return 1;
  if (n % 2 == 1 && n > 9) return calc(n - 9) + 1;
  return -inf;
}
int main() {
  scanf("%d", &t);
  while (t--) scanf("%d", &n), printf("%d\n", max(calc(n), -1));
  return 0;
}

```