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;
const long long mod = 1e9 + 7;
const long long N = 1e6;
const long long inf = 1e18;
long long add(long long a, long long b) {
  long long res = (a + b) % mod;
  return (res < 0) ? res + mod : res;
}
long long mul(long long a, long long b) {
  long long res = (a * 1LL * b) % mod;
  return (res < 0) ? res + mod : res;
}
bool compare(long long x, long long y) { return x > y; }
struct Compare {
  bool operator()(vector<long long> A, vector<long long> B) {}
};
long long prime[11];
void SieveOfEratosthenes() {
  prime[0] = prime[1] = 1;
  for (long long p = 2; p * p <= 11; p++) {
    if (prime[p] == 0) {
      for (long long i = p * p; i <= 11; i += p) {
        prime[i] = 1;
      }
    }
  }
}
void solve() {
  SieveOfEratosthenes();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    if (n == 1 || (n <= 11 && prime[n] == 0)) {
      cout << -1;
    } else if (n == 13) {
      cout << 2;
    } else {
      long long ans = n / 4;
      long long rem = n % 4;
      if (rem == 1 || rem == 3) ans--;
      cout << ans;
    }
    cout << "\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
