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 = 1000000007;
const long long MAX = 100005;
const long double PI = 3.14159265359;
const long double G = 9.807;
const long long INF = 1e18;
const long double EPS = 1e-6;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
bool isprime(long long a) {
  if (a == 2) return 1;
  if (!(a & 1)) return 0;
  for (long long i = 3; i * i <= a; i += 2)
    if (a % i == 0) return 0;
  return 1;
}
long long mpow(long long base, long long exponent, long long modulus) {
  if (modulus == 1) return 0;
  long long result = 1;
  base = base % modulus;
  while (exponent) {
    if (exponent % 2 == 1) result = (result * base) % modulus;
    exponent = exponent >> 1;
    base = (base * base) % modulus;
  }
  return result;
}
long long minv(long long a, long long mod) {
  long long _gcd = gcd(a, mod);
  assert(_gcd == 1);
  return mpow(a, mod - 2, mod);
}
int q;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> q;
  while (q--) {
    int n;
    cin >> n;
    int fr = n / 4;
    int rem = n % 4;
    if (rem == 0) {
      cout << fr;
    } else if (rem == 1) {
      if (fr >= 2) {
        cout << fr - 1;
      } else {
        cout << -1;
      }
    } else if (rem == 2) {
      if (fr >= 1) {
        cout << fr;
      } else {
        cout << -1;
      }
    } else if (rem == 3) {
      if (fr >= 3) {
        cout << fr - 1;
      } else {
        cout << -1;
      }
    }
    cout << '\n';
  }
  return 0;
}
