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;
void print(vector<pair<int, int>> v) {
  for (auto x : v) cout << x.first << " " << x.second << endl;
}
int n, k;
int dp[5000][5000];
vector<int> v(5000);
int f(int idx, int times) {
  if (idx == n) {
    if (times <= k)
      return 0;
    else
      return -(int)2e9;
  }
  if (dp[idx][times] != -1) return dp[idx][times];
  int next_team_pos =
      upper_bound(v.begin(), v.begin() + n, v[idx] + 5) - v.begin();
  return dp[idx][times] = max(f(next_team_pos, times + 1) + next_team_pos - idx,
                              f(idx + 1, times));
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> v[i];
  sort(v.begin(), v.begin() + n);
  memset(dp, -1, sizeof(dp));
  cout << f(0, 0) << endl;
  return 0;
}