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, M, K, T;
int students[5050];
int minTeamIndices[5050];
int dp[5050][5050] = {0};
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &students[i]);
  }
  sort(students + 1, students + N + 1);
  minTeamIndices[0] = 1;
  minTeamIndices[1] = 1;
  for (int i = 2; i <= N; i++) {
    int x = students[i];
    int j = minTeamIndices[i - 1];
    while (students[j] < x - 5) j++;
    minTeamIndices[i] = j;
  }
  for (int n = 1; n <= N; n++) {
    for (int i = 0; i < K + 1; i++) {
      int dontMakeTeam = dp[n - 1][i];
      int teamStartIndex = minTeamIndices[n];
      int makeTeam = dp[teamStartIndex - 1][i - 1] + n - teamStartIndex + 1;
      if (i == 0) makeTeam = 0;
      dp[n][i] = (((dontMakeTeam) > (makeTeam)) ? (dontMakeTeam) : (makeTeam));
    }
  }
  printf("%d", dp[N][K]);
  return 0;
}