#include <bits/stdc++.h>
#pragma optimize("O3")
using namespace std;
const long long MOD = 998244353;
const long long INF = 1e18 + 7;
const int base = 2e5 + 1;
const long long MAX = 1e6;
const double EPS = 1e-9;
const double PI = acos(-1.);
const int MAXN = 1e6 + 147;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  int q;
  cin >> q;
  while (q--) {
    int n;
    cin >> n;
    int x = n / 4;
    int y = (n - 4 * x) / 6;
    int z = (n - 4 * x - 6 * y) / 9;
    int mx = -1;
    for (int i = (max(0, x - 10)); i < (x + 10); ++i) {
      for (int j = (max(0, y - 10)); j < (y + 10); ++j) {
        for (int k = (max(0, z - 10)); k < (z + 10); ++k) {
          if (4 * i + 6 * j + 9 * k == n) {
            mx = max(mx, i + j + k);
          }
        }
      }
    }
    printf("%d\n", mx);
  }
  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>
#pragma optimize("O3")
using namespace std;
const long long MOD = 998244353;
const long long INF = 1e18 + 7;
const int base = 2e5 + 1;
const long long MAX = 1e6;
const double EPS = 1e-9;
const double PI = acos(-1.);
const int MAXN = 1e6 + 147;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  int q;
  cin >> q;
  while (q--) {
    int n;
    cin >> n;
    int x = n / 4;
    int y = (n - 4 * x) / 6;
    int z = (n - 4 * x - 6 * y) / 9;
    int mx = -1;
    for (int i = (max(0, x - 10)); i < (x + 10); ++i) {
      for (int j = (max(0, y - 10)); j < (y + 10); ++j) {
        for (int k = (max(0, z - 10)); k < (z + 10); ++k) {
          if (4 * i + 6 * j + 9 * k == n) {
            mx = max(mx, i + j + k);
          }
        }
      }
    }
    printf("%d\n", mx);
  }
  return 0;
}

```