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>
namespace fast_IO {
char buf[10000005];
int cur_p = -1, len;
inline int getch() {
  if (++cur_p >= len) {
    len = fread(buf, 1, 10000005, stdin);
    cur_p = 0;
  }
  return buf[cur_p];
}
inline long long getll() {
  long long res = 0;
  int c = getch(), f = 1;
  if (c == '-') {
    f = -1;
    c = getch();
  }
  for (; c < 48 || c > 57; c = getch())
    ;
  for (; c >= 48 && c <= 57; c = getch())
    res = (res << 3) + (res << 1) + c - 48;
  return f == 1 ? res : -res;
}
inline int getint() {
  int res = 0, c = getch(), f = 1;
  if (c == '-') {
    f = -1;
    c = getch();
  }
  for (; c < 48 || c > 57; c = getch())
    ;
  for (; c >= 48 && c <= 57; c = getch())
    res = (res << 3) + (res << 1) + c - 48;
  return f == 1 ? res : -res;
}
}  // namespace fast_IO
using fast_IO::getch;
using fast_IO::getint;
using fast_IO::getll;
using namespace std;
int main() {
  int q;
  scanf("%d", &q);
  while (q--) {
    int n;
    scanf("%d", &n);
    if ((n & 3) == 1)
      printf("%d\n", (n >> 2) < 2 ? -1 : (n >> 2) - 1);
    else if ((n & 3) == 2)
      printf("%d\n", !(n >> 2) ? -1 : (n >> 2));
    else if ((n & 3) == 3)
      printf("%d\n", (n >> 2) < 3 ? -1 : (n >> 2) - 1);
    else
      printf("%d\n", n >> 2);
  }
  return 0;
}
