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 n, k;
vector<int> in(5000);
int tb[5000][5000];
int dp(int i, int count) {
  if (i == n) {
    if (count <= k)
      return 0;
    else
      return -100000;
  }
  if (tb[i][count] != -1) return tb[i][count];
  int hi = upper_bound(in.begin(), in.begin() + n, in[i] + 5) - in.begin();
  int resp = dp(hi, count + 1) + hi - i;
  resp = max(resp, dp(i + 1, count));
  return tb[i][count] = resp;
}
int main() {
  memset(tb, -1, sizeof(tb));
  cin >> n >> k;
  for (int i = 0; i < n; ++i) cin >> in[i];
  sort(in.begin(), in.begin() + n);
  cout << dp(0, 0) << endl;
  return 0;
}