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 i, j, x, y, z, m, n, k, l, r, f, mi = 1e9;
int a[200];
bool pr(int n) {
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return true;
  return false;
}
int main() {
  scanf("%d", &n);
  a[1] = a[2] = a[3] = -1;
  a[4] = 1;
  for (int i = 5; i < 100; i++) {
    int t = -1;
    if (pr(i)) t = 1;
    for (int j = 4; j <= i / 2; j++)
      if (pr(j) && pr(i - j)) t = max(t, a[j] + a[i - j]);
    a[i] = t;
  }
  while (n--) {
    scanf("%d", &m);
    int ans;
    if (m < 20)
      ans = a[m];
    else {
      ans = m / 4;
      m = m % 4;
      if (m == 1) ans--;
      if (m == 3) ans--;
    }
    printf("%d\n", ans);
  }
}
