#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int SIZE = 1e6 + 10;
int dp[SIZE];
int main() {
  vector<int> fac = {4, 6, 8, 9, 10, 12, 14};
  memset((dp), -1, sizeof((dp)));
  dp[0] = 0;
  for (int i = (1); i < (SIZE); ++i) {
    for (int j = 0; j < (((int)(fac).size())); ++j) {
      if (fac[j] <= i && dp[i - fac[j]] != -1) {
        if (dp[i] == -1 || dp[i] < dp[i - fac[j]] + 1) {
          dp[i] = dp[i - fac[j]] + 1;
        }
      }
    }
  }
  int(Q);
  scanf("%d", &Q);
  for (int cs = 0; cs < (Q); ++cs) {
    int(x);
    scanf("%d", &x);
    if (x < SIZE)
      printf("%d\n", dp[x]);
    else {
      int v = x / 4 - 10000;
      printf("%d\n", v + dp[x - v * 4]);
    }
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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>
using namespace std;
const int MOD = 1e9 + 7;
const int SIZE = 1e6 + 10;
int dp[SIZE];
int main() {
  vector<int> fac = {4, 6, 8, 9, 10, 12, 14};
  memset((dp), -1, sizeof((dp)));
  dp[0] = 0;
  for (int i = (1); i < (SIZE); ++i) {
    for (int j = 0; j < (((int)(fac).size())); ++j) {
      if (fac[j] <= i && dp[i - fac[j]] != -1) {
        if (dp[i] == -1 || dp[i] < dp[i - fac[j]] + 1) {
          dp[i] = dp[i - fac[j]] + 1;
        }
      }
    }
  }
  int(Q);
  scanf("%d", &Q);
  for (int cs = 0; cs < (Q); ++cs) {
    int(x);
    scanf("%d", &x);
    if (x < SIZE)
      printf("%d\n", dp[x]);
    else {
      int v = x / 4 - 10000;
      printf("%d\n", v + dp[x - v * 4]);
    }
  }
  return 0;
}

```