#include <bits/stdc++.h>
using namespace std;
template <class T>
void splitstr(const string &s, vector<T> &out) {
  istringstream in(s);
  out.clear();
  copy(istream_iterator<T>(in), istream_iterator<T>(), back_inserter(out));
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
static void redirect(int argc, const char **argv) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  if (argc > 1) {
    static filebuf f;
    f.open(argv[1], ios::in);
    cin.rdbuf(&f);
    if (!cin) {
      cerr << "Failed to open '" << argv[1] << "'" << endl;
      exit(1);
    }
  }
  if (argc > 2) {
    static filebuf f;
    f.open(argv[2], ios::out | ios::trunc);
    cout.rdbuf(&f);
    if (!cout) {
      cerr << "Failed to open '" << argv[2] << "'" << endl;
    }
  }
}
int main(int argc, const char **argv) {
  redirect(argc, argv);
  vector<int> comp{4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20};
  int M = 1000;
  vector<int> dp(M, -1);
  dp[0] = 0;
  for (int i = 1; i < M; i++) {
    for (int v : comp)
      if (v <= i && dp[i - v] != -1) dp[i] = max(dp[i], dp[i - v] + 1);
  }
  for (int i = M / 2; i < M; i++) assert(dp[i] == dp[i - 4] + 1);
  int Q;
  cin >> Q;
  for (int i = 0; i < Q; i++) {
    int v;
    cin >> v;
    if (v < M)
      cout << dp[v] << '\n';
    else {
      int r = (v - M) / 4 + 1;
      cout << dp[v - 4 * r] + r << '\n';
    }
  }
  return 0;
}

### Prompt

Generate 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;
template <class T>
void splitstr(const string &s, vector<T> &out) {
  istringstream in(s);
  out.clear();
  copy(istream_iterator<T>(in), istream_iterator<T>(), back_inserter(out));
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
static void redirect(int argc, const char **argv) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  if (argc > 1) {
    static filebuf f;
    f.open(argv[1], ios::in);
    cin.rdbuf(&f);
    if (!cin) {
      cerr << "Failed to open '" << argv[1] << "'" << endl;
      exit(1);
    }
  }
  if (argc > 2) {
    static filebuf f;
    f.open(argv[2], ios::out | ios::trunc);
    cout.rdbuf(&f);
    if (!cout) {
      cerr << "Failed to open '" << argv[2] << "'" << endl;
    }
  }
}
int main(int argc, const char **argv) {
  redirect(argc, argv);
  vector<int> comp{4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20};
  int M = 1000;
  vector<int> dp(M, -1);
  dp[0] = 0;
  for (int i = 1; i < M; i++) {
    for (int v : comp)
      if (v <= i && dp[i - v] != -1) dp[i] = max(dp[i], dp[i - v] + 1);
  }
  for (int i = M / 2; i < M; i++) assert(dp[i] == dp[i - 4] + 1);
  int Q;
  cin >> Q;
  for (int i = 0; i < Q; i++) {
    int v;
    cin >> v;
    if (v < M)
      cout << dp[v] << '\n';
    else {
      int r = (v - M) / 4 + 1;
      cout << dp[v - 4 * r] + r << '\n';
    }
  }
  return 0;
}

```