#include <bits/stdc++.h>
using namespace std;
int ans[105];
int n;
bool comp(int x) {
  for (int i = 2; i < x; i++)
    if (x % i == 0) return true;
  return false;
}
int main() {
  for (int i = 1; i < 105; i++) {
    ans[i] = -1;
    for (int j = 2; j <= i; j++)
      if (comp(j) && ans[i - j] != -1) ans[i] = max(ans[i], ans[i - j] + 1);
  }
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    if (x < 105)
      printf("%d\n", ans[x]);
    else
      printf("%d\n", ans[96 + (x % 4)] + (x / 4 - 96 / 4));
  }
  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;
int ans[105];
int n;
bool comp(int x) {
  for (int i = 2; i < x; i++)
    if (x % i == 0) return true;
  return false;
}
int main() {
  for (int i = 1; i < 105; i++) {
    ans[i] = -1;
    for (int j = 2; j <= i; j++)
      if (comp(j) && ans[i - j] != -1) ans[i] = max(ans[i], ans[i - j] + 1);
  }
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    if (x < 105)
      printf("%d\n", ans[x]);
    else
      printf("%d\n", ans[96 + (x % 4)] + (x / 4 - 96 / 4));
  }
  return 0;
}

```