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>
using namespace std;
int mpow(int base, int exp);
void ipgraph(long long n, long long m);
void dfs(int u, int par);
const int mod = 1000000007;
const int N = 3e5, M = N;
vector<long long> g[N];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  long long t, i, j, k, p, q, r, x, y, u, v, n, m;
  cin >> t;
  while (t--) {
    cin >> n;
    if ((n < 4) || (n == 5) || (n == 7) || (n == 11)) {
      printf("-1\n");
      continue;
    }
    if (n % 4 == 0) {
      printf("%d\n", n / 4);
      continue;
    }
    if (n % 4 == 2) {
      printf("%d\n", n / 4);
      continue;
    }
    if (n % 4 == 1) {
      printf("%d\n", (n / 4) - 1);
      continue;
    }
    if (n % 4 == 3) {
      printf("%d\n", (n / 4) - 1);
      continue;
    }
  }
  return 0;
}
int mpow(int base, int exp) {
  base %= mod;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((long long)result * base) % mod;
    base = ((long long)base * base) % mod;
    exp >>= 1;
  }
  return result;
}
void ipgraph(long long n, long long m) {
  int i, u, v;
  while (m--) {
    cin >> u >> v;
    g[u - 1].push_back(v - 1);
    g[v - 1].push_back(u - 1);
  }
}
void dfs(int u, int par) {
  for (int v : g[u]) {
    if (v == par) continue;
    dfs(v, u);
  }
}
