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.*;
public class Main{
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int k=sc.nextInt();
        int[] A=new int[n];
        for(int i=0;i<n;i++)A[i]=sc.nextInt();
        Arrays.sort(A);
        int[] dp=new int[n];
        for(int i=0;i<k;i++){
            int[] ndp=new int[n];
            for(int j=0,r=0;r<n;){
                while(r<n&&A[j]+5>=A[r]){//以r为结尾 尽可能取得多
                    ndp[r]=Math.max(r==0?0:ndp[r-1],r-j+1+(j==0?0:dp[j-1]));
                    r++;
                }
                while(r<n&&A[j]+5<A[r]){
                    j++;
                }
                if(r<n) {
                   ndp[r]=Math.max(r==0?0:ndp[r-1],r-j+1+(j==0?0:dp[j-1]));
                }
            }
            dp=ndp;
        }
        System.out.println(dp[n-1]);
    }
}