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>
const int N = 8000;
using namespace std;
int d[N];
int main() {
  ios_base::sync_with_stdio(0);
  vector<int> v;
  for (int i = 1; i < N; i++) {
    bool good = false;
    for (int j = 2; j < i; j++) {
      if (i % j == 0) good = true;
    }
    if (good) v.push_back(i);
  }
  for (int i = 1; i < N; i++) {
    d[i] = -1;
    for (int j : v) {
      if (i >= j && d[i - j] != -1) d[i] = max(d[i], d[i - j] + 1);
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int x;
    cin >> x;
    if (x < N) {
      cout << d[x] << "\n";
    } else {
      int nx = x % (N / 2);
      nx += N / 2;
      assert(d[nx] != -1);
      cout << d[nx] + (x - nx) / 4 << "\n";
    }
  }
}
