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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
bool isprime(int x) {
  if (x == 0) return false;
  if (x == 1) return true;
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q = 0;
  cin >> q;
  while (q--) {
    int n;
    cin >> n;
    int ans = -1;
    if (n % 4 == 0) ans = n / 4;
    if ((n - 6) >= 0 && (n - 6) % 4 == 0) ans = max(ans, (n - 6) / 4 + 1);
    if ((n - 9) >= 0 && (n - 9) % 4 == 0) ans = max(ans, (n - 9) / 4 + 1);
    if ((n - 15) >= 0 && (n - 15) % 4 == 0) ans = max(ans, (n - 15) / 4 + 2);
    cout << ans << endl;
  }
  return 0;
}
