#include <bits/stdc++.h>
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2,ssse3,tune=native")
using namespace std;
using ld = long double;
using vec = complex<ld>;
void solve() {
  int a;
  cin >> a;
  if (a < 4 || a == 5 || a == 7 || a == 11) {
    cout << "-1\n";
    return;
  }
  if (a % 4 == 0) {
    cout << a / 4 << "\n";
  } else if (a % 4 == 1) {
    cout << a / 4 - 1 << "\n";
  } else if (a % 4 == 2) {
    cout << a / 4 << "\n";
  } else if (a % 4 == 3) {
    cout << a / 4 - 1 << "\n";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2,ssse3,tune=native")
using namespace std;
using ld = long double;
using vec = complex<ld>;
void solve() {
  int a;
  cin >> a;
  if (a < 4 || a == 5 || a == 7 || a == 11) {
    cout << "-1\n";
    return;
  }
  if (a % 4 == 0) {
    cout << a / 4 << "\n";
  } else if (a % 4 == 1) {
    cout << a / 4 - 1 << "\n";
  } else if (a % 4 == 2) {
    cout << a / 4 << "\n";
  } else if (a % 4 == 3) {
    cout << a / 4 - 1 << "\n";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

```