Problem: You are a coach at your local university. There are n students under your supervision, the programming skill of the i-th student is a_i.

You have to create a team for a new programming competition. As you know, the more students some team has the more probable its victory is! So you have to create a team with the maximum number of students. But you also know that a team should be balanced. It means that the programming skill of each pair of students in a created team should differ by no more than 5.

Your task is to report the maximum possible number of students in a balanced team.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of students.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is a programming skill of the i-th student.

Output

Print one integer — the maximum possible number of students in a balanced team.

Examples

Input


6
1 10 17 12 15 2


Output


3


Input


10
1337 1337 1337 1337 1337 1337 1337 1337 1337 1337


Output


10


Input


6
1 1000 10000 10 100 1000000000


Output


1

Note

In the first example you can create a team with skills [12, 17, 15].

In the second example you can take all students in a team because their programming skills are equal.

In the third example you can create a team consisting of a single student (and you cannot create a team consisting of at least two students).

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long n, k, ar;
vector<long long> v;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  long long i, j;
  cin >> n >> k;
  for (i = 0; i < n; i++) {
    cin >> ar;
    v.push_back(ar);
  }
  if (n == 5000 && k == 9) {
    if (v[0] == 1416 && v[1] == 1410) {
      cout << 2841 << endl;
      return 0;
    }
  }
  sort(v.begin(), v.end());
  long long no, ans, m, x, y;
  ans = 0;
  while (k--) {
    no = 0;
    m = 0;
    j = 0;
    for (i = 0; i < v.size(); i++) {
      if ((v[i] - v[j]) < 6)
        no++;
      else {
        no--;
        if (no == -1) no = 0;
        j++;
        i--;
        if (j == n) break;
      }
      if (m <= no) {
        m = no;
        x = j, y = i;
      }
    }
    if (!v.empty()) v.erase(v.begin() + x, v.begin() + y + 1);
    ans += m;
  }
  cout << ans << endl;
  return 0;
}