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 pi = acos(-1.0);
void fast() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long int comb(long long int m, long long int n) {
  long long int mat[100][100];
  long long int i, j;
  if (n > m) return 0;
  if ((n == 0) || (m == n)) return 1;
  for (j = 0; j < n; j++) {
    mat[0][j] = 1;
    if (j == 0) {
      for (i = 1; i <= m - n; i++) mat[i][j] = i + 1;
    } else {
      for (i = 1; i <= m - n; i++) mat[i][j] = mat[i - 1][j] + mat[i][j - 1];
    }
  }
  return (mat[m - n][n - 1]);
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int ans = 0;
    while (n > 15) {
      ans++;
      n = n - 4;
    }
    if (n == 4 || n == 6 || n == 9) {
      ans++;
    }
    if (n == 8 || n == 13 || n == 10 || n == 15) {
      ans = ans + 2;
    }
    if (n == 12 || n == 14) {
      ans = ans + 3;
    }
    if (ans == 0) {
      cout << -1 << '\n';
    } else {
      cout << ans << '\n';
    }
  }
}
