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 INF = 9223372036854775807;
const long long INFINT = 2147483647;
const int N = 1000006;
const int NN = 1006;
const int MOD = 1000000007;
void DIE(string s) {
  cout << s << '\n';
  exit(0);
}
void DIE(char x) {
  cout << x << '\n';
  exit(0);
}
void DIE(long long x) {
  cout << x << '\n';
  exit(0);
}
void DIE(int x) {
  cout << x << '\n';
  exit(0);
}
void YES() { DIE("YES"); }
void Yes() { DIE("Yes"); }
void yes() { DIE("yes"); }
void NO() { DIE("NO"); }
void No() { DIE("No"); }
void no() { DIE("no"); }
int a[11] = {0, -1, -1, -1, 1, -1, 1, -1, 2, 1, 2};
int solve(int x) {
  int ans = 0;
  if (x % 4 == 0) return x / 4;
  if (x & 1) {
    x -= 9;
    ans++;
  }
  if (x < 0) return -1;
  if (x <= 10) {
    if (a[x] == -1) return -1;
    return a[x] + ans;
  }
  x -= 10;
  ans += x / 4;
  x %= 4;
  x += 10;
  while (x > 10) {
    ans++;
    x -= 4;
  }
  if (a[x] == -1) return -1;
  return ans + a[x];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    cout << solve(x) << '\n';
  }
}
