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>
long long z = 1000000000000000001;
using namespace std;
vector<bool> vec(100001, 1);
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
bool sortbysecdesc(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second > b.second;
}
int egcd(int a, int b) {
  if (b == 0)
    return a;
  else
    return egcd(b, a % b);
}
double logn(double a, double base) { return log2(a) / log2(base); }
bool isprime(long long n) {
  if (n == 1) return false;
  bool b = true;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      b = false;
      break;
    }
  }
  return b;
}
void sieve() {
  int n = 100001;
  for (int i = 2; i * i <= n; i++) {
    if (vec[i]) {
      for (int j = i * i; j <= n; j += i) vec[j] = false;
    }
  }
}
struct node {
  int p, cf, cb;
  node *next;
};
bool compare(double a, double b) {
  double eps = 0.000001;
  return fabs(a - b) <= eps;
}
long long count_Divisors(long long n) {
  sieve();
  long long cnt = 0, ans = 1;
  if (n == 1) {
    return 1;
  }
  for (long long i = 2; (i * i * i) <= n; i++) {
    if (n % i == 0) {
      int cnt = 1;
      while (n > 1 && n % i == 0) {
        n /= i;
        cnt++;
      }
      ans *= cnt;
    }
  }
  if (isprime(n))
    ans *= 2;
  else if ((floor(sqrt(n)) == ceil(sqrt(n))) && isprime((int)sqrt(n)))
    ans *= 3;
  else if (n != 1)
    ans *= 4;
  return ans;
}
void solve() {
  int a[16];
  a[1] = a[2] = a[3] = a[5] = a[7] = a[11] = -1;
  a[4] = 1;
  a[6] = 1;
  a[8] = 2;
  a[9] = 1;
  a[10] = 2;
  a[12] = 3;
  a[13] = 2;
  a[14] = 3;
  a[15] = 2;
  int q, n;
  cin >> q;
  while (q--) {
    cin >> n;
    if (n <= 15)
      cout << a[n] << '\n';
    else {
      if (n % 4 == 0)
        cout << n / 4 << '\n';
      else if (n % 4 == 1)
        cout << 1 + (n - 9) / 4 << '\n';
      else if (n % 4 == 2)
        cout << 1 + (n - 6) / 4 << '\n';
      else
        cout << 2 + (n - 15) / 4 << '\n';
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}
