When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?

You're given a sequence of n data points a1, ..., an. There aren't any big jumps between consecutive data points — for each 1 ≤ i < n, it's guaranteed that |ai + 1 - ai| ≤ 1.

A range [l, r] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let M be the maximum and m the minimum value of ai for l ≤ i ≤ r; the range [l, r] is almost constant if M - m ≤ 1.

Find the length of the longest almost constant range.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of data points.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000).

Output

Print a single number — the maximum length of an almost constant range of the given sequence.

Examples

Input

5
1 2 3 3 2


Output

4


Input

11
5 4 5 5 6 7 8 8 8 7 6


Output

5

Note

In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.

In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].

import java.io.*;
import java.net.Socket;
import java.security.spec.ECField;
import java.util.HashMap;
import java.util.StringTokenizer;
import static java.util.Arrays.sort;


public class Main {

    static int a[];
    static int l[];
    static int arr[];
    static HashMap<Integer, Boolean> map = new HashMap<>();

    public static void main (String [] args) throws Exception {
        InputReader reader = new InputReader();
        PrintWriter out = new PrintWriter(System.out);

        int n, last=0,ans=0;

        n = reader.nextInt();
        arr=new int[n];
        int []step=new int[n];

        for(int i =0 ; i < n ; i++)
        {
            arr[i]=reader.nextInt();

        }
        for(int i =0 ; i < n-1; i++)
        {
            step[i]= arr[i]-arr[i+1];

        }


        int temp = 0 ,old=0;
        for(int i =0 ; i < n-1; i++)
        {
          //System.out.print ( step[i]+" ");




            if(step[i] != 0 && step[i] == last ) // two con
            {
                ans=Math.max(temp , ans);
                temp=0;
                temp= i - old;
            }else{
                temp++;
                ans=Math.max(ans,temp);

            }

            if(step[i] != 0 ){
                old=i;
                last= step[i];
            }



        }

        System.out.println(ans+1);


        out.flush();
    }



    static class InputReader {
        BufferedReader reader;
        StringTokenizer tokenizer;

        InputReader() {
            reader = new BufferedReader(new InputStreamReader(System.in));
            tokenizer = null;
        }

        String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return tokenizer.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }
    }
}
