#include <bits/stdc++.h>
using namespace std;
const int maxn = 1010;
int f[maxn];
int is[maxn];
void prepare() {
  for (int i = 2; i <= 1000; i++)
    if (!is[i]) {
      for (int j = i * 2; j <= 1000; j += i) is[j] = true;
    }
  for (int i = 1; i <= 1000; i++) {
    if (is[i])
      f[i] = 1;
    else
      f[i] = -1;
    for (int j = 4; j <= i - 4; j++)
      if (f[i - j] != -1 && f[j] != -1) f[i] = max(f[i], f[i - j] + f[j]);
  }
}
int n, x;
int main() {
  scanf("%d", &n);
  prepare();
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    int ans = 0;
    if (x > 1000) {
      ans += (x - 1000) / 4;
      x -= ans * 4;
      while (x > 1000) x -= 4, ans++;
    }
    ans += f[x];
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Generate 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;
const int maxn = 1010;
int f[maxn];
int is[maxn];
void prepare() {
  for (int i = 2; i <= 1000; i++)
    if (!is[i]) {
      for (int j = i * 2; j <= 1000; j += i) is[j] = true;
    }
  for (int i = 1; i <= 1000; i++) {
    if (is[i])
      f[i] = 1;
    else
      f[i] = -1;
    for (int j = 4; j <= i - 4; j++)
      if (f[i - j] != -1 && f[j] != -1) f[i] = max(f[i], f[i - j] + f[j]);
  }
}
int n, x;
int main() {
  scanf("%d", &n);
  prepare();
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    int ans = 0;
    if (x > 1000) {
      ans += (x - 1000) / 4;
      x -= ans * 4;
      while (x > 1000) x -= 4, ans++;
    }
    ans += f[x];
    printf("%d\n", ans);
  }
  return 0;
}

```