//     Author: Rodney Shaghoulian
//     Github: github.com/RodneyShag
// HackerRank: hackerrank.com/RodneyShag

import java.util.Scanner;

// Algorithm: 
//  1 - Notice there is a special restriction on the range of values in the array: values are between 0 and 100.
//  2 - Since the order of elements in our array does not matter, we can just keep the counts of each number.
//  3 - Create a new 100-element "counts" array to store the counts of each number.
//  4 - In our "counts" array, the adjacent pair with the largest sum is our final solution.
//  5 - This is because the original numbers represented by the 2 adjacent slots in counts[] can be used to create a 
//      maximal solution

// Time complexity: O(n)
public class Solution {
    static final int MAX_VALUE = 100;
    
    static int solve(int[] array) {
        /* Save counts of each number */
        int[] counts = new int[MAX_VALUE];
        for (int num : array) {
            counts[num]++;
        }
        
        /* Find adjacent pair with largest sum */
        int max = 2;
        for (int i = 1; i < 100; i++) {
            int adjacentSum = counts[i] + counts[i-1];
            max = Math.max(max, adjacentSum);
        }
        return max;
    }

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int[] array = new int[n];
        for (int i = 0; i < n; i++) {
            array[i] = scan.nextInt();
        }
        scan.close();
        
        System.out.println(solve(array));
    }
}