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;
bool sortbysecdesc(const pair<long long, long long> &a,
                   const pair<long long, long long> &b) {
  if (a.first == b.first)
    return a.second > b.second;
  else
    return 0;
}
const long long N = 100010;
bool prime[N];
void SieveOfEratosthenes() {
  long long n = N - 3;
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
long long i, j;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  SieveOfEratosthenes();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    if (n == 1 || n == 2 || n == 3 || n == 5 || n == 7 || n == 11) {
      cout << "-1" << '\n';
    } else if (n % 4 == 0) {
      cout << n / 4 << '\n';
    } else if (n == 6) {
      cout << 1 << '\n';
    } else {
      long long temp = n / 4;
      long long x = n - temp * 4;
      if (x == 1) {
        cout << 1 + (n - 9) / 4 << '\n';
      } else if (x == 2) {
        cout << 1 + (n - 6) / 4 << '\n';
      } else if (x == 3) {
        cout << 2 + (n - 15) / 4 << '\n';
      }
    }
  }
  return 0;
}
