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;
using ll = long long int;
int dp[5001][5001];
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (auto &e : a) cin >> e;
  sort(a.begin(), a.end());
  vector<int> sz(n);
  for (int i = 0; i < n; ++i) {
    auto it = upper_bound(a.begin(), a.end(), a[i] + 5);
    sz[i] = (it - a.begin()) - i;
  }
  fill(&dp[0][0], &dp[5001][0], -1);
  dp[0][0] = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < k; ++j) {
      if (dp[i][j] == -1) break;
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      dp[i + sz[i]][j + 1] = max(dp[i + sz[i]][j + 1], dp[i][j] + sz[i]);
    }
    if (dp[i][k] != -1) {
      dp[i + 1][k] = max(dp[i + 1][k], dp[i][k]);
    }
  }
  int ans = *max_element(&dp[n][1], &dp[n][k + 1]);
  cout << ans << endl;
  return 0;
}