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 bs(vector<long long>& a, long long target) {
  long long n = a.size();
  long long l = 1;
  long long r = n - 1;
  long long answer = -1;
  while (l <= r) {
    long long mid = (l + (r - l) / 2);
    if (a[mid] <= target) {
      answer = mid;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  return answer;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> a(n + 1, 0);
  vector<long long> prefix(n + 1, 0);
  for (long long i = 1; i <= n; i++) cin >> a[i];
  sort(a.begin() + 1, a.end());
  long long sum = 0;
  for (long long i = 1; i <= n; i++) {
    sum += a[i];
    prefix[i] = sum;
  }
  long long maxsum = 0;
  long long answer = 0;
  for (long long i = 1; i <= n; i++) {
    long long curindex = i;
    long long maximum = a[i] + 5;
    long long nextindex = bs(a, maximum);
    long long total = nextindex - curindex + 1;
    if (total > answer) {
      answer = total;
    }
  }
  cout << answer << "\n";
  return 0;
}