#include <bits/stdc++.h>
using namespace std;
void getint(int &v) {
  char ch, fu = 0;
  for (ch = '*'; (ch < '0' || ch > '9') && ch != '-'; ch = getchar())
    ;
  if (ch == '-') fu = 1, ch = getchar();
  for (v = 0; ch >= '0' && ch <= '9'; ch = getchar()) v = v * 10 + ch - '0';
  if (fu) v = -v;
}
int T, n, ans;
int main() {
  scanf("%d", &T);
  while (T--) {
    getint(n);
    if (n % 2 == 0) {
      if (n == 2)
        ans = -1;
      else if (n % 4 == 0)
        ans = n / 4;
      else
        ans = n / 4;
    } else {
      if (n < 9)
        ans = -1;
      else {
        ans = 1;
        n -= 9;
        if (n == 2)
          ans = -1;
        else if (n % 4 == 0)
          ans += n / 4;
        else
          ans += n / 4;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Your challenge is to write 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 getint(int &v) {
  char ch, fu = 0;
  for (ch = '*'; (ch < '0' || ch > '9') && ch != '-'; ch = getchar())
    ;
  if (ch == '-') fu = 1, ch = getchar();
  for (v = 0; ch >= '0' && ch <= '9'; ch = getchar()) v = v * 10 + ch - '0';
  if (fu) v = -v;
}
int T, n, ans;
int main() {
  scanf("%d", &T);
  while (T--) {
    getint(n);
    if (n % 2 == 0) {
      if (n == 2)
        ans = -1;
      else if (n % 4 == 0)
        ans = n / 4;
      else
        ans = n / 4;
    } else {
      if (n < 9)
        ans = -1;
      else {
        ans = 1;
        n -= 9;
        if (n == 2)
          ans = -1;
        else if (n % 4 == 0)
          ans += n / 4;
        else
          ans += n / 4;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}

```