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 long long MOD = 1000000007;
const int INF = 1e9;
const long long IINF = 1e18;
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const char dir[4] = {'D', 'R', 'U', 'L'};
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &x : v) is >> x;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (int i = 0; i < v.size(); ++i) {
    os << v[i] << (i + 1 == v.size() ? "" : " ");
  }
  return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  os << '(' << p.first << ',' << p.second << ')';
  return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const map<T, U> &m) {
  os << '{';
  for (auto itr = m.begin(); itr != m.end(); ++itr) {
    os << '(' << itr->first << ',' << itr->second << ')';
    if (++itr != m.end()) os << ',';
    --itr;
  }
  os << '}';
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &s) {
  os << '{';
  for (auto itr = s.begin(); itr != s.end(); ++itr) {
    os << *itr;
    if (++itr != s.end()) os << ',';
    --itr;
  }
  os << '}';
  return os;
}
void debug_out() { cerr << '\n'; }
template <class Head, class... Tail>
void debug_out(Head &&head, Tail &&...tail) {
  cerr << head;
  if (sizeof...(Tail) > 0) cerr << ", ";
  debug_out(move(tail)...);
}
template <typename T>
T gcd(T x, T y) {
  return y != 0 ? gcd(y, x % y) : x;
}
template <typename T>
T lcm(T x, T y) {
  return x / gcd(x, y) * y;
}
template <class T1, class T2>
inline bool chmin(T1 &a, T2 b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T1, class T2>
inline bool chmax(T1 &a, T2 b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
vector<int> ng{1, 2, 3, 5, 7, 11};
void solve() {
  int n;
  cin >> n;
  for (int x : ng) {
    if (x == n) {
      cout << -1 << '\n';
      return;
    }
  }
  int ans;
  if (n % 4 == 0) {
    ans = n / 4;
  } else if (n % 4 == 1) {
    ans = 1 + (n - 9) / 4;
  } else if (n % 4 == 2) {
    ans = 1 + (n - 6) / 4;
  } else {
    ans = 2 + (n - 15) / 4;
  }
  cout << ans << '\n';
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int q;
  cin >> q;
  for (; q--;) {
    solve();
  }
}
