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.InputStream;
import java.io.InputStreamReader;
import java.util.*;

public class Question{

    static int ans = -1;


    public static void main(String[] args) throws IOException {
        Reader.init(System.in);
        int n = Reader.nextInt();
        int[] arr = new int[n];
        for (int i = 0 ; i < n ; i++){
            arr[i] = Reader.nextInt();
        }
        int ans = -1;
        MergeSort ob = new MergeSort();
        ob.sort(arr,0,n-1);
        for (int i = 0 ; i < n ; i++){
            //System.out.println(binarySearchCount(arr,n,arr[i]+5));
            ans = Math.max(binarySearchCount(arr,n,arr[i]+5)- i,ans);
        }

        System.out.println(ans);



    }

    static int binarySearchCount(int arr[], int n, int key)
    {
        int left = 0, right = n;

        int mid = 0;
        while (left < right) {
            mid = (right + left) >> 1;

            // Check if key is present in array
            if (arr[mid] == key) {

                // If duplicates are present it returns
                // the position of last element
                while (mid+1<n && arr[mid + 1] == key)
                    mid++;
                break;
            }

            // If key is smaller, ignore right half
            else if (arr[mid] > key)
                right = mid;

                // If key is greater, ignore left half
            else
                left = mid + 1;
        }

        // If key is not found in array then it will be
        // before mid
        while (mid > - 1 && arr[mid] > key)
            mid--;

        // Return mid + 1 because of 0-based indexing
        // of array
        return mid + 1;
    }
}



class Reader {
    static BufferedReader reader;
    static StringTokenizer tokenizer;

    /** call this method to initialize reader for InputStream */
    static void init(InputStream input) {
        reader = new BufferedReader(
                new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }

    /** get next word */
    static String next() throws IOException {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                    reader.readLine() );
        }
        return tokenizer.nextToken();
    }

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

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

class MergeSort {
    // Merges two subarrays of arr[].
    // First subarray is arr[l..m]
    // Second subarray is arr[m+1..r]
    void merge(int arr[], int l, int m, int r) {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;

        /* Create temp arrays */
        int L[] = new int[n1];
        int R[] = new int[n2];

        /*Copy data to temp arrays*/
        for (int i = 0; i < n1; ++i)
            L[i] = arr[l + i];
        for (int j = 0; j < n2; ++j)
            R[j] = arr[m + 1 + j];


        /* Merge the temp arrays */

        // Initial indexes of first and second subarrays
        int i = 0, j = 0;

        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) {
                arr[k] = L[i];
                i++;
            } else {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

        /* Copy remaining elements of L[] if any */
        while (i < n1) {
            arr[k] = L[i];
            i++;
            k++;
        }

        /* Copy remaining elements of R[] if any */
        while (j < n2) {
            arr[k] = R[j];
            j++;
            k++;
        }
    }

    // Main function that sorts arr[l..r] using
    // merge()
    void sort(int arr[], int l, int r) {
        if (l < r) {
            // Find the middle point
            int m = (l + r) / 2;

            // Sort first and second halves
            sort(arr, l, m);
            sort(arr, m + 1, r);

            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

    /* A utility function to print array of size n */
    static void printArray(int arr[]) {
        int n = arr.length;
        for (int i = 0; i < n; ++i)
            System.out.print(arr[i] + " ");
        System.out.println();
    }
}