#include <bits/stdc++.h>
static std::vector<char> is_prime;
static void one_case(void) {
  static std::vector<long long> factors;
  long long n;
  std::scanf("%lld", &n);
  if (n <= 11 && (is_prime[n] || n == 1)) {
    std::printf("-1\n");
  } else {
    std::printf("%lld\n", (n % 4) % 2 == 0 ? n / 4 : n / 4 - 1);
  }
}
int main(void) {
  is_prime.assign(14, true);
  is_prime[0] = is_prime[1] = false;
  for (long long i = 2; i < is_prime.size(); ++i) {
    if (!is_prime[i]) continue;
    for (long long j = i + i; j < is_prime.size(); j += i) is_prime[j] = false;
  }
  long long t;
  std::scanf("%lld", &t);
  while (t--) one_case();
  return 0;
}

### Prompt

Please formulate 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>
static std::vector<char> is_prime;
static void one_case(void) {
  static std::vector<long long> factors;
  long long n;
  std::scanf("%lld", &n);
  if (n <= 11 && (is_prime[n] || n == 1)) {
    std::printf("-1\n");
  } else {
    std::printf("%lld\n", (n % 4) % 2 == 0 ? n / 4 : n / 4 - 1);
  }
}
int main(void) {
  is_prime.assign(14, true);
  is_prime[0] = is_prime[1] = false;
  for (long long i = 2; i < is_prime.size(); ++i) {
    if (!is_prime[i]) continue;
    for (long long j = i + i; j < is_prime.size(); j += i) is_prime[j] = false;
  }
  long long t;
  std::scanf("%lld", &t);
  while (t--) one_case();
  return 0;
}

```