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;
const double PI = acos(0) * 2;
const double EPS = 1e-8;
const long long MOD = 1e9 + 7;
const int MAXN = 1e5 + 5;
const int oo = 1e9;
const double foo = 1e30;
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcounll(s);
}
template <class T>
T sqr(T x) {
  return x * x;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
int f[MAXN];
int isPrime(int u) {
  if (u == 1) return 1;
  for (int i = 2; i * i <= u; i++) {
    if (u % i == 0) return 0;
  }
  return 1;
}
int main() {
  for (int i = 1; i <= 100; i++) {
    f[i] = -1;
    if (!isPrime(i)) f[i] = i;
    for (int j = 1; j < i; j++) {
      if (isPrime(j) || isPrime(i - j)) continue;
      f[i] = j;
    }
  }
  int T, n;
  cin >> T;
  while (T--) {
    cin >> n;
    if (n <= 11 && f[n] == -1) {
      cout << -1;
    } else {
      if (n % 4 == 0) {
        cout << n / 4;
      }
      if (n % 4 == 2) {
        cout << n / 4;
      }
      if (n % 4 == 1) {
        cout << n / 4 - 1;
      }
      if (n % 4 == 3) {
        cout << n / 4 - 1;
      }
    }
    cout << '\n';
  }
  return 0;
}
