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;
vector<int> V;
int DP[5010][5010];
int DP2[5010];
int n, k;
int solve(const int idx, const int rem) {
  if (idx == n) return 0;
  if (rem <= 0) return 0;
  if (DP[idx][rem] != -1) return DP[idx][rem];
  auto it = lower_bound(V.begin(), V.end(), V[idx] + 6);
  int tot;
  if (DP2[idx] != -1)
    tot = DP2[idx];
  else
    DP2[idx] = tot = it - V.begin() - idx;
  return DP[idx][rem] =
             max(solve(idx + 1, rem), tot + solve(it - V.begin(), rem - 1));
}
int main() {
  int a;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a), V.push_back(a);
  sort(V.begin(), V.end());
  memset(DP, -1, sizeof(DP));
  memset(DP2, -1, sizeof(DP2));
  printf("%d\n", solve(0, k));
  return 0;
}