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;
int Q, n, ans[20];
int main() {
  ans[1] = -1;
  ans[2] = -1;
  ans[3] = -1;
  ans[4] = 1;
  ans[5] = -1;
  ans[6] = 1;
  ans[7] = -1;
  ans[8] = 2;
  ans[9] = 1;
  ans[10] = 2;
  ans[11] = -1;
  ans[12] = 3;
  ans[13] = 2;
  ans[14] = 3;
  ans[15] = 2;
  scanf("%d", &Q);
  for (int i = 1; i <= Q; i++) {
    scanf("%d", &n);
    if (n <= 15)
      printf("%d\n", ans[n]);
    else {
      if (n % 4 == 0)
        printf("%d\n", n / 4);
      else if (n % 4 == 1)
        printf("%d\n", (n - 9) / 4 + 1);
      else if (n % 4 == 2)
        printf("%d\n", (n - 6) / 4 + 1);
      else if (n % 4 == 3)
        printf("%d\n", (n - 15) / 4 + 2);
    }
  }
  return 0;
}
