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.
#include <bits/stdc++.h>
using namespace std;
template <class T>
T sqr(const T &a) {
  return a * a;
}
const int MAX = 10000;
int d[3 * MAX];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 0; i < (2 * MAX); i++) {
    if (i == 0 || d[i]) {
      d[i + 4] = max(d[i + 4], d[i] + 1);
      d[i + 6] = max(d[i + 6], d[i] + 1);
      d[i + 9] = max(d[i + 9], d[i] + 1);
    }
  }
  int T;
  scanf("%d", &T);
  for (int i = 0; i < (T); i++) {
    int n;
    scanf("%d", &n);
    int ans = 0;
    if (n > MAX) {
      ans = (n - MAX) / 4;
      n -= ans * 4;
    }
    ans += d[n];
    if (ans == 0)
      puts("-1");
    else
      printf("%d\n", ans);
  }
}
