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 fx[] = {-1, 1, 0, 0};
int fy[] = {0, 0, -1, 1};
int dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
int kx[] = {1, 1, -1, -1, 2, 2, -2, -2};
int ky[] = {2, -2, 2, -2, 1, -1, 1, -1};
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
  return a.first < b.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> ara;
  for (int i = 1; i <= n; i++) {
    int d;
    cin >> d;
    ara.push_back(d);
  }
  sort(ara.begin(), ara.end());
  int maxi = 0;
  for (int i = 0; i < n; i++) {
    int v = ara[i] + 5;
    vector<int>::iterator it = upper_bound(ara.begin(), ara.end(), v);
    int cur = (it - ara.begin()) - i;
    maxi = max(maxi, cur);
  }
  cout << maxi << endl;
  return 0;
}