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 long long modn = 1000000007;
inline long long mod(long long x) { return x % modn; }
template <class num>
inline void rd(num &x) {
  char c;
  while (isspace(c = getchar()))
    ;
  bool neg = false;
  if (!isdigit(c))
    neg = (c == '-'), x = 0;
  else
    x = c - '0';
  while (isdigit(c = getchar())) x = (x << 3) + (x << 1) + c - '0';
  if (neg) x = -x;
}
int n;
const int N = 1123;
int cp[N];
int memo[N];
int solve(int n) {
  if (n == 0) return 0;
  if (n < 0) return -100000;
  int &r = memo[n];
  if (r != -1) return r;
  r = -2;
  for (int i = 4; i <= n; i++)
    if (cp[i] && solve(n - i) >= 0) r = max(r, 1 + solve(n - i));
  return r;
}
int a[N];
int main() {
  memset(memo, -1, sizeof memo);
  int i, j;
  rd(n);
  for (i = 2; i < N; i++)
    if (!cp[i])
      for (j = i + i; j < N; j += i) cp[j] = 1;
  for (i = 1; i <= 200; i++) a[i] = solve(i);
  for (i = 0; i < n; i++) {
    int x;
    rd(x);
    int mx = -1;
    for (j = (x % 4); j <= x && j <= 200; j += 4)
      if (a[j] >= 0) mx = max(mx, (x - j) / 4 + a[j]);
    printf("%d\n", mx);
  }
}
