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;
void maximiza(int &a, int b) { a = max(a, b); }
const int limite = 1000;
const int infinito = 1000000000;
int esprimo[limite];
int solucion[limite];
int main() {
  for (int i = 2; i < limite; i++) esprimo[i] = 1;
  for (int i = 2; i < limite; i++)
    if (esprimo[i])
      for (int j = i + i; j < limite; j += i) esprimo[j] = 0;
  for (int i = 1; i < limite; i++) solucion[i] = -1;
  for (int i = 4; i < limite; i++) {
    if (not esprimo[i]) solucion[i] = 1;
    for (int j = 2; j < i - 1; j++) {
      if (not esprimo[j]) {
        int anterior = i - j;
        if (not esprimo[anterior]) maximiza(solucion[i], 2);
        if (solucion[anterior] > 0)
          maximiza(solucion[i], 1 + solucion[anterior]);
      }
    }
  }
  int casos;
  cin >> casos;
  for (int cas = 0; cas < casos; cas++) {
    int x;
    cin >> x;
    if (x < limite)
      cout << solucion[x] << endl;
    else {
      int resta = (x - limite / 2) / 4 * 4;
      cout << resta / 4 + solucion[x - resta] << endl;
    }
  }
}
