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].
//package com.company;
import java.math.BigInteger;
import java.util.*;
import java.io.*;

public class Main  {

    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        int n=in.nextInt();
        int A[]=new int[n];
        int firsti=0;
        for(int i=0;i<n;i++)
        {
            A[i]=in.nextInt();
        }
        for(int i=0;i<n;i++)
        {
            if(A[i]!=A[0])
            {
                break;
            }
            firsti++;
        }
        if(firsti==n)
        {
            System.out.println(n);
        }
        else
        {
            int prevsize=firsti;
            int max=firsti;
            int cursize=0;
            int cur=A[firsti];
            int prev=A[firsti-1];
            int previ=0;
            int now=A[firsti-1];
            while (true)
            {
                //System.out.println(max);
                while (firsti<n&&(A[firsti]==cur||A[firsti]==prev))
                {
                    if(now!=A[firsti])
                    {
                        now=A[firsti];
                        previ=firsti;
                    }
                    cursize++;
                    firsti++;
                }
                if(firsti==n)
                {
                    max=Math.max(max,cursize+prevsize);
                    break;
                }
                else
                {
                    prev=A[firsti-1];
                    cur=A[firsti];
                    max=Math.max(max,cursize+prevsize);
                    prevsize=firsti-previ;
                    cursize=0;

                }
                //System.out.println(firsti+" "+previ);
            }
            System.out.println(max);

        }

    }

}
