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.Scanner;

public class Main {

    public static void main(String[] args) {

        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int m = scanner.nextInt();

        int[] a = new int[n + 1];
        for (int i = 1; i <= n; i++)
            a[i] = scanner.nextInt();
        quickSort(a, 1, n);
        int[][] result = new int[n + 1][m + 1];
        int[] oneTeamCount = new int[n + 1];
        int last = 1;
        oneTeamCount[1] = 1;
        result[1][1] = 1;
        for (int i = 2; i <= n; i++) {
            while (a[i] - a[last] > 5)
                last++;
            oneTeamCount[i] = i - last + 1;
            result[i][1] = Math.max(result[i - 1][1], oneTeamCount[i]);
        }
        int temp;
        for (int j = 2; j <= m; j++)
            for (int i = j; i <= n; i++) {
                temp = Math.max(j - 1, i - oneTeamCount[i]);
                result[i][j] = Math.max(result[i - 1][j], result[temp][j - 1] + i - temp);
            }
        System.out.println(result[n][m]);

    }

    public static void quickSort(int[] a, int left, int right) {

        int mid = a[(left + right) / 2];
        int x = left, y = right, temp;
        while (x <= y) {
            while (x <= right && a[x] < mid) x++;
            while (y >= left && a[y] > mid) y--;
            if (x <= y) {
                temp = a[x];
                a[x] = a[y];
                a[y] = temp;
                x++;
                y--;
            }
        }

        if (left < y)
            quickSort(a, left, y);
        if (x < right)
            quickSort(a, x, right);
    }



}