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.*;
import java.io.*;

public class E544 {
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int n = sc.nextInt(); int k = sc.nextInt();
        long [] a = new long[n + 1];
        for (int i = 1; i <= n; i++) a[i] = sc.nextInt();
        Arrays.sort(a, 1, n + 1);
        PriorityQueue<Pair> pq = new PriorityQueue<>(Comparator.comparingLong(x -> -x.val));
        int [] last = new int[n + 1];
        for (int i = n; i >= 1; i--) {
            while (!pq.isEmpty() && pq.peek().val > 5 + a[i]) {
                Pair p = pq.poll();
                last[p.i] = i + 1;
            }
            pq.add(new Pair(i, a[i]));
        }
        for (int i = 1; i <= n; i++) if (last[i] == 0) last[i] = 1;
        int [][] dp = new int[n + 1][k + 1];
        long ans = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= k; j++) {
                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j]);
                dp[i][j] = Math.max(dp[Math.max(0, last[i] - 1)][j - 1] + (i - last[i] + 1), dp[i][j]);
                ans = Math.max(ans, dp[i][j]);
            }
        }
        out.println(ans);
        out.close();
    }

    static class Pair {
        int i; long val;
        Pair(int i, long val) {
            this.i = i; this.val = val;
        }
    }


    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }


    }

}