#include <bits/stdc++.h>
using namespace std;
int q, n;
bool ip(int n) {
  int i;
  if (n == 1) {
    return 1;
  }
  for (i = 2; i * i <= n; i++) {
    if (!(n % i)) {
      return 0;
    }
  }
  return 1;
}
int main() {
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &n);
    if (n % 4 == 3) {
      if (n < 15) {
        printf("-1\n");
      } else {
        printf("%d\n", (n - 15) / 4 + 2);
      }
      continue;
    }
    int ans = n / 4;
    n %= 4;
    if (!n) {
      printf("%d\n", ans);
      continue;
    }
    int t = 0;
    while (ip(n)) {
      n += 4;
      ans--;
      t = 1;
    }
    if (ans < 0) {
      printf("-1\n");
    } else {
      printf("%d\n", ans + t);
    }
  }
  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;
int q, n;
bool ip(int n) {
  int i;
  if (n == 1) {
    return 1;
  }
  for (i = 2; i * i <= n; i++) {
    if (!(n % i)) {
      return 0;
    }
  }
  return 1;
}
int main() {
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &n);
    if (n % 4 == 3) {
      if (n < 15) {
        printf("-1\n");
      } else {
        printf("%d\n", (n - 15) / 4 + 2);
      }
      continue;
    }
    int ans = n / 4;
    n %= 4;
    if (!n) {
      printf("%d\n", ans);
      continue;
    }
    int t = 0;
    while (ip(n)) {
      n += 4;
      ans--;
      t = 1;
    }
    if (ans < 0) {
      printf("-1\n");
    } else {
      printf("%d\n", ans + t);
    }
  }
  return 0;
}

```