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.io.*;
import java.util.*;

public class CF1133E {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		int k = Integer.parseInt(st.nextToken());
		st = new StringTokenizer(br.readLine());
		Integer[] aa = new Integer[n];
		for (int i = 0; i < n; i++)
			aa[i] = Integer.parseInt(st.nextToken());
		Arrays.sort(aa);
		int[] pp = new int[n];
		for (int i = 0, j = 0; j < n; j++) {
			while (aa[i] < aa[j] - 5)
				i++;
			pp[j] = i - 1;
		}
		int[] dp = new int[n];
		while (k-- > 0) {
			for (int j = n - 1; j >= 0; j--) {
				int i = pp[j];
				dp[j] = Math.max(dp[j], (i >= 0 ? dp[i] : 0) + j - i);
			}
			for (int j = 1; j < n; j++)
				dp[j] = Math.max(dp[j], dp[j - 1]);
		}
		System.out.println(dp[n - 1]);
	}
}