#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:250000000")
using namespace std;
int dp[1001];
int a[] = {4, 6, 9, 10};
int go(int n) {
  if (n < 0) return -(int)1e9;
  if (n == 0) return 0;
  int& res = dp[n];
  if (res != -1) return res;
  res = -(int)1e9;
  for (int i = 0; i < 4; ++i) res = max(res, go(n - a[i]) + 1);
  return res;
}
int main() {
  memset(dp, -1, sizeof(dp));
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; ++i) {
    int n;
    scanf("%d", &n);
    int k = 0;
    if (n >= 900) {
      k += (n - 900) / 4;
      n -= 4 * k;
    }
    int res = go(n);
    if (res < 0)
      printf("-1\n");
    else
      printf("%d\n", res + k);
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:250000000")
using namespace std;
int dp[1001];
int a[] = {4, 6, 9, 10};
int go(int n) {
  if (n < 0) return -(int)1e9;
  if (n == 0) return 0;
  int& res = dp[n];
  if (res != -1) return res;
  res = -(int)1e9;
  for (int i = 0; i < 4; ++i) res = max(res, go(n - a[i]) + 1);
  return res;
}
int main() {
  memset(dp, -1, sizeof(dp));
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; ++i) {
    int n;
    scanf("%d", &n);
    int k = 0;
    if (n >= 900) {
      k += (n - 900) / 4;
      n -= 4 * k;
    }
    int res = go(n);
    if (res < 0)
      printf("-1\n");
    else
      printf("%d\n", res + k);
  }
  return 0;
}

```