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;
int main() {
  int n, res = 0;
  cin >> n;
  vector<int> v;
  vector<pair<int, int> > vv;
  for (int i = 0; i < n; i++) {
    int c;
    cin >> c;
    v.push_back(c);
  }
  sort(v.begin(), v.end());
  int cnt = 1;
  for (int i = 1; i < n; i++) {
    if (v[i] != v[i - 1]) {
      vv.push_back(make_pair(v[i - 1], cnt));
      cnt = 1;
    } else
      cnt++;
  }
  vv.push_back(make_pair(v[n - 1], cnt));
  for (int i = 0; i < vv.size(); i++) {
    int temp = vv[i].first, j = i + 1;
    cnt = vv[i].second;
    while ((vv[j].first - temp) <= 5 && j < vv.size()) {
      cnt += vv[j].second;
      j++;
    }
    res = max(res, cnt);
  }
  cout << res << endl;
}