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;
long long const mod = 1000000007;
long long const md = 998244353;
long long mypow(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
ifstream in("txt.in");
ofstream out("txt.out");
int f[20];
int main() {
  cout << fixed << setprecision(12);
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  long long t;
  cin >> t;
  f[1] = -1;
  f[3] = -1;
  f[5] = -1;
  f[8] = 2;
  f[9] = 1;
  f[11] = -1;
  f[13] = 2;
  f[16] = 4;
  f[2] = -1;
  f[4] = 1;
  f[6] = 1;
  f[7] = -1;
  f[10] = 2;
  f[12] = 3;
  f[14] = 3;
  f[15] = 2;
  f[17] = 3;
  while (t--) {
    long long x;
    cin >> x;
    if (x <= 17)
      cout << f[x] << '\n';
    else {
      long long u = x / 4;
      long long k = x % 4;
      while (f[k] <= 0) {
        --u;
        k += 4;
      }
      cout << u + f[k] << '\n';
    }
  }
  return 0;
}
