#include <bits/stdc++.h>
int main() {
  int t, d;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &d);
    if (d < 4)
      printf("-1\n");
    else if (d == 5)
      printf("-1\n");
    else if (d == 7)
      printf("-1\n");
    else if (d == 11)
      printf("-1\n");
    else {
      if (d % 2 == 1) {
        d -= 9;
        int p = d / 4;
        int q = d % 4;
        while (q % 6 != 0) {
          p--;
          q += 4;
        }
        printf("%d\n", p + 1 + q / 6);
      } else {
        int p = d / 4;
        int q = d % 4;
        while (q % 6 != 0) {
          p--;
          q += 4;
        }
        printf("%d\n", p + q / 6);
      }
    }
  }
  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>
int main() {
  int t, d;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &d);
    if (d < 4)
      printf("-1\n");
    else if (d == 5)
      printf("-1\n");
    else if (d == 7)
      printf("-1\n");
    else if (d == 11)
      printf("-1\n");
    else {
      if (d % 2 == 1) {
        d -= 9;
        int p = d / 4;
        int q = d % 4;
        while (q % 6 != 0) {
          p--;
          q += 4;
        }
        printf("%d\n", p + 1 + q / 6);
      } else {
        int p = d / 4;
        int q = d % 4;
        while (q % 6 != 0) {
          p--;
          q += 4;
        }
        printf("%d\n", p + q / 6);
      }
    }
  }
  return 0;
}

```