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;
vector<int> gr[200005];
bool visited[200005];
void dfs(int node, int parent) {
  for (auto it : gr[node]) {
    if (it != parent) dfs(it, node);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int a[16];
    a[0] = -1;
    a[1] = -1;
    a[2] = -1;
    a[3] = -1;
    a[4] = 1;
    a[5] = -1;
    a[6] = 1;
    a[7] = -1;
    a[8] = 2;
    a[9] = 1;
    a[10] = 2;
    a[11] = -1;
    a[12] = 3;
    a[13] = 2;
    a[14] = 3;
    a[15] = 2;
    if (n <= 15) {
      cout << a[n] << endl;
      continue;
    }
    int x = n, a1 = 0;
    bool ok = true;
    while (1) {
      if (x <= 15) {
        if (a[x] == -1) {
          ok = false;
          break;
        } else {
          a1 += a[x];
          break;
        }
      }
      x -= 4;
      a1++;
    }
    if (ok && x >= 1) {
      cout << a1 << endl;
      continue;
    }
    ok = true;
    a1 = 0;
    x = n;
    while (1) {
      if (x <= 15) {
        if (a[x] == -1) {
          ok = false;
          break;
        } else {
          a1 += a[x];
          break;
        }
      }
      x -= 6;
      a1++;
    }
    if (ok && x >= 1) {
      cout << a1 << endl;
      continue;
    }
    ok = true;
    a1 = 0;
    x = n;
    while (1) {
      if (x <= 15) {
        if (a[x] == -1) {
          ok = false;
          break;
        } else {
          a1 += a[x];
          break;
        }
      }
      x -= 9;
      a1++;
    }
    if (ok && x >= 1) {
      cout << a1 << endl;
      continue;
    }
    cout << -1 << endl;
  }
  return 0;
}
