#include <bits/stdc++.h>
using namespace std;
template <typename T, typename T1>
T amax(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
const int32_t mod = 1e9 + 7;
const long long INF = 1e18;
const long long N = 1e5 + 5, M = 103;
void solve() {
  long long n;
  cin >> n;
  if (n < 4) {
    cout << "-1\n";
    return;
  }
  if (n % 2 == 0) {
    cout << n / 4 << "\n";
    return;
  }
  n -= 9;
  if (n > 3 || n == 0)
    cout << n / 4 + 1 << "\n";
  else
    cout << "-1\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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>
using namespace std;
template <typename T, typename T1>
T amax(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
const int32_t mod = 1e9 + 7;
const long long INF = 1e18;
const long long N = 1e5 + 5, M = 103;
void solve() {
  long long n;
  cin >> n;
  if (n < 4) {
    cout << "-1\n";
    return;
  }
  if (n % 2 == 0) {
    cout << n / 4 << "\n";
    return;
  }
  n -= 9;
  if (n > 3 || n == 0)
    cout << n / 4 + 1 << "\n";
  else
    cout << "-1\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```