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.Scanner;

public class E {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(), k = sc.nextInt();
        int[] a = new int[5005], Prev = new int[5005], maxd = new int[5005];
        int[][] f = new int[5005][5005];
        for (int i = 1; i <= n; i++)
            a[i] = sc.nextInt();
        Arrays.sort(a,1,n+1);
        int l = 1, res = 0;
        for (int r = 1; r <= n; r++) {
            while (a[r] - a[l] > 5)
                l++;
            Prev[r] = l;
        }
        for (int t = 1; t <= k; t++) {
            for (int i = 1; i <= n; i++) {
                f[i][t] = maxd[Prev[i] - 1] + (i - Prev[i] + 1);
                if (t == k)
                    res = Math.max(res, f[i][t]);
            }
            for (int i = 1; i <= n; i++)
                maxd[i] = Math.max(maxd[i - 1], f[i][t]);
        }
        System.out.println(res);
    }
}