#include <bits/stdc++.h>
using namespace std;
using ll = long long;
bool isPrime(int x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
int Q, N;
int dp[101];
int main() {
  dp[1] = dp[2] = dp[3] = -1;
  for (int i = 4; i <= 100; i++) {
    dp[i] = -1;
    for (int j = 4; j <= i; j++) {
      if (!isPrime(j) && dp[i - j] != -1) {
        dp[i] = max(dp[i], 1 + dp[i - j]);
      }
    }
  }
  cin >> Q;
  for (int q = 0, _n = Q; q < _n; q++) {
    cin >> N;
    int res = -1;
    for (int i = 0; i <= 100 && i <= N; i++) {
      int rem = N - i;
      if (rem % 4 == 0 && dp[i] != -1) {
        res = max(res, dp[i] + rem / 4);
      }
    }
    cout << res << endl;
  }
}

### 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;
using ll = long long;
bool isPrime(int x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
int Q, N;
int dp[101];
int main() {
  dp[1] = dp[2] = dp[3] = -1;
  for (int i = 4; i <= 100; i++) {
    dp[i] = -1;
    for (int j = 4; j <= i; j++) {
      if (!isPrime(j) && dp[i - j] != -1) {
        dp[i] = max(dp[i], 1 + dp[i - j]);
      }
    }
  }
  cin >> Q;
  for (int q = 0, _n = Q; q < _n; q++) {
    cin >> N;
    int res = -1;
    for (int i = 0; i <= 100 && i <= N; i++) {
      int rem = N - i;
      if (rem % 4 == 0 && dp[i] != -1) {
        res = max(res, dp[i] + rem / 4);
      }
    }
    cout << res << endl;
  }
}

```