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, ans;
int dp[5005][5005], a[5005];
int brute(int pos, int k) {
  if (pos == n || k == 0) return 0;
  if (dp[pos][k] != -1) return dp[pos][k];
  int ans = 0;
  ans = brute(pos + 1, k);
  int ind;
  for (ind = pos + 1; ind < n; ind++) {
    if (a[ind] - a[pos] > 5) break;
  }
  ans = max(ans, ind - pos + brute(ind, k - 1));
  dp[pos][k] = ans;
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  memset(dp, -1, sizeof(dp));
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  int ans = brute(0, k);
  cout << ans;
  cerr << "Time elapsed : " << 1.0 * clock() / CLOCKS_PER_SEC << " sec \n";
}