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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;

public class a {
	public static void main(String[] args) throws IOException {
		StreamTokenizer st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		st.nextToken();
		int n = (int) st.nval;
		Integer[] a = new Integer[n];
		Map<Long, Long> map = new HashMap<>();
		for (int i = 0; i < n; i++) {
			st.nextToken();
			a[i] = (int) st.nval;
		}
		Arrays.sort(a);
		Integer[] b = new Integer[n];
		int k = 2, j = 1;
		int min = a[0];
		b[0] = 1;
		int max = 1;
		for (int i = 1; i < n; i++) {
			while (i < n && a[i] - min <= 5) {
				b[i] = k++;
				i++;
			}
			max = Math.max(max, b[i - 1]);
			while (i < n && a[i] - min > 5) {
				min = a[j++];
				k--;
			}
			i--;
		}
		System.out.println(max);
	}
}