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 double Eps = 1e-9;
int Mod = 1000000000 + 7;
int n, m, q, k;
int Arr[100010];
bool prime[10010];
bool ok = true;
void sieve() {
  for (int i = 2; i * i <= 1e4; i++) {
    if (prime[i] == true)
      for (int j = i * 2; j <= 1e4; j++) {
        prime[j] = false;
      }
  }
}
int dp[100010];
int cnt;
int main() {
  ios::sync_with_stdio(0);
  ;
  memset(prime, true, sizeof(prime));
  sieve();
  cin >> n;
  for (int i = 0; i < n; i++) {
    cnt = 0;
    cin >> Arr[i];
    if (Arr[i] & 1 && Arr[i] >= 9) {
      Arr[i] -= 9;
      cnt++;
    }
    if (Arr[i] % 4 == 0 && Arr[i] >= 0) {
      cout << cnt + Arr[i] / 4 << endl;
      continue;
    } else if ((Arr[i] - 6) % 4 == 0 && (Arr[i] - 6) >= 0) {
      cout << cnt + 1 + (Arr[i] - 6) / 4 << endl;
      continue;
    }
    cout << -1 << endl;
  }
}
