#include <bits/stdc++.h>
using namespace std;
const int maxn = 155555;
inline int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= ch == '-', ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return f ? -x : x;
}
int t, n;
int main() {
  t = read();
  while (t--) {
    n = read();
    int ans = 0;
    if (n & 1) {
      if (n < 9) {
        puts("-1");
        continue;
      }
      if (n == 9) {
        puts("1");
        continue;
      }
      n -= 9;
      ans++;
    }
    if (n <= 3) {
      puts("-1");
      continue;
    }
    printf("%d\n", (n >> 2) + ans);
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 = 155555;
inline int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= ch == '-', ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return f ? -x : x;
}
int t, n;
int main() {
  t = read();
  while (t--) {
    n = read();
    int ans = 0;
    if (n & 1) {
      if (n < 9) {
        puts("-1");
        continue;
      }
      if (n == 9) {
        puts("1");
        continue;
      }
      n -= 9;
      ans++;
    }
    if (n <= 3) {
      puts("-1");
      continue;
    }
    printf("%d\n", (n >> 2) + ans);
  }
}

```