#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int INF = 1000000000;
const ll LINF = 1001002003004005006ll;
int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
struct IOSetup {
  IOSetup() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout << fixed << setprecision(12);
  }
} iosetup;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (int i = 0; i < (int)v.size(); i++)
    os << v[i] << (i + 1 == (int)v.size() ? "" : " ");
  return os;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &x : v) is >> x;
  return is;
}
void solve() {
  ll n;
  cin >> n;
  ll res;
  if (n % 4 == 0) res = n / 4;
  if (n % 4 == 1) res = (n / 4) - 1;
  if (n % 4 == 2) res = (n / 4);
  if (n % 4 == 3) res = (n / 4) - 1;
  if (res <= 0) res = -1;
  if (n == 11) res = -1;
  cout << res << endl;
}
signed main() {
  int q;
  cin >> q;
  while (q--) solve();
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to 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;
using ll = long long;
const int INF = 1000000000;
const ll LINF = 1001002003004005006ll;
int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
struct IOSetup {
  IOSetup() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout << fixed << setprecision(12);
  }
} iosetup;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (int i = 0; i < (int)v.size(); i++)
    os << v[i] << (i + 1 == (int)v.size() ? "" : " ");
  return os;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &x : v) is >> x;
  return is;
}
void solve() {
  ll n;
  cin >> n;
  ll res;
  if (n % 4 == 0) res = n / 4;
  if (n % 4 == 1) res = (n / 4) - 1;
  if (n % 4 == 2) res = (n / 4);
  if (n % 4 == 3) res = (n / 4) - 1;
  if (res <= 0) res = -1;
  if (n == 11) res = -1;
  cout << res << endl;
}
signed main() {
  int q;
  cin >> q;
  while (q--) solve();
  return 0;
}

```