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>
std::mt19937_64 mersenne_twister(
    std::chrono::steady_clock::now().time_since_epoch().count());
namespace McDicCP {
bool precomputed = false;
std::vector<int> DP;
const int plimit = 15000;
std::vector<bool> is_this_prime(plimit, true);
std::vector<int> primes, composites;
void precompute() {
  if (precomputed) return;
  for (int i = 2; i < plimit; i++) {
    if (is_this_prime[i]) {
      primes.push_back(i);
      for (int j = 2 * i; j < plimit; j += i) is_this_prime[j] = false;
    } else
      composites.push_back(i);
  }
  DP.clear();
  DP.resize(plimit, -1);
  for (int i = 4; i < plimit; i++) {
    if (!is_this_prime[i]) DP[i] = 1;
    for (int j : composites) {
      if (j > 20 || i - j < 0)
        break;
      else if (DP[i - j] == -1)
        continue;
      DP[i] = std::max(DP[i], DP[i - j] + 1);
    }
  }
  precomputed = true;
}
int solve() {
  precompute();
  int n;
  std::cin >> n;
  if (n < plimit)
    return DP[n];
  else {
    int offset = (n - plimit) / 4 + 1;
    n = n - 4 * offset;
    return DP[n] + offset;
  }
}
}  // namespace McDicCP
int main(int argc, char **argv) {
  int testcases = 1;
  std::cin >> testcases;
  for (int t = 1; t <= testcases; t++) std::cout << McDicCP::solve() << '\n';
  return 0;
}
