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;
long long int dp[199999];
long long int func(long long int x, long long int *dp) {
  if (x > 100000 && x / 100000 != 1) {
    return func(100000, dp) * (x / 100000 - 1) + func(x % 100000 + 100000, dp);
  } else {
    if (dp[x] != 0 && x != 0)
      return dp[x];
    else if (dp[x] == 0 && x == 0)
      return dp[x];
    else if (x & 1) {
      dp[x] = 1 + func(x - 9, dp);
      return (1 + func(x - 9, dp));
    } else {
      dp[x] = 1 + func(x - 4, dp);
      return (1 + func(x - 4, dp));
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int q;
  cin >> q;
  memset(dp, 0, sizeof(dp));
  dp[1] = -1;
  dp[2] = -1;
  dp[3] = -1;
  dp[4] = 1;
  dp[5] = -1;
  dp[6] = 1;
  dp[7] = -1;
  dp[8] = 2;
  dp[9] = 1;
  dp[10] = 2;
  dp[11] = -1;
  while (q--) {
    long long int n;
    cin >> n;
    cout << func(n, dp) << '\n';
  }
  return 0;
}
