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 JAVA):

Solution:
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public class E {
    private static int binSearchEnd(int[] a, int k) {
        //int n = a.size();
        int n = a.length;

        int start = 0;
        int end = n - 1;

        while (start <= end) {
            int middle = start + (end - start) / 2;

            int val = a[middle];

            if (val <= k) {
                start = middle + 1;
            } else {
                end = middle - 1;
            }
        }

        return start - 1;
    }

    private static int binSearchBeginning(List<Integer> a, int k) {
        int n = a.size();

        int start = 0;
        int end = n - 1;

        while (start <= end) {
            int middle = start + (end - start) / 2;

            int val = a.get(middle);

            if (val < k) {
                start = middle + 1;
            } else {
                end = middle - 1;
            }
        }

        return end + 1;
    }

    private static int solveRec(int[][] memo, int[] ratings, int i, int k) {
        //int n = ratings.size();
        int n = ratings.length;

        if (k <= 0) {
            return 0;
        }

        if (i >= n) {
            return 0;
        }

        if (i == n - 1) {
            return 1;
        }

        if (memo[i][k] != 0) {
            return memo[i][k];
        }

        int rating = ratings[i];

        //int intervalStart = binSearchBeginning(ratings, rating + 5);
        //int intervalStart = i;
        int intervalEnd = binSearchEnd(ratings, rating + 5);

        int res = Math.max(solveRec(memo, ratings, i + 1, k), solveRec(memo, ratings, intervalEnd + 1, k - 1) + (intervalEnd - i) + 1);

        //for (int j = intervalStart; j <= intervalEnd; j++) {
        //    res = Math.max(res, solveRec(memo, ratings, j + 1, k - 1) + (j - intervalStart + 1));
        //}

        memo[i][k] = res;

        //System.out.printf("Starting from %d with %d left, result is %d%n", i, k, res);

        return res;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int k = sc.nextInt();
        //List<Integer> ratings = new ArrayList<>(n);
        int[] ratings = new int[n];

        for (int i = 0; i < n; i++) {
            ratings[i] = sc.nextInt();
            //ratings.add(sc.nextInt());
        }

        //ratings.sort(Integer::compareTo);
        Arrays.sort(ratings);

        int[][] memo = new int[n][k + 1];

        System.out.println(solveRec(memo, ratings, 0, k));
    }
}