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.util.*;
import java.lang.System;
public class Solution{
	static PrintStream out=System.out;
	static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    static StringTokenizer st;
	static Scanner sc=new Scanner(System.in);
	
	
	static long n,arr[];
	static void Task(){
		MultiSet ms=new MultiSet();
		n=ni();arr=nla(i(n));
		long[] left=new long[i(n)];
		long j=0;
		for(int i=0;i<n;i++){
			while(j<n){
				ms.add(arr[i(j)]);
				//out.println(j);
				long min=ms.first();
				long max=ms.last();
				//out.println(min+" "+max);
				if(max-min>1 ){
					//out.println("in if");
					left[i]=j;
					ms.remove(arr[i(i)]);
					ms.remove(arr[i(j)]);
					break;
				}
				else j+=1;
			}
			if(j>=n)left[i]=n;
		}
		//for(int i=0;i<n;i++)out.print(left[i]+" " );
		long ans=-1;
		for(int i=0;i<n;i++){
			ans=Math.max(ans,left[i]-i);
		}
		out.println(ans);
	}







	public static void main(String args[]){
		Task();
	}
	static int ni(){
		return Integer.parseInt(ns());
	}
	static long nl(){
		 return Long.parseLong(ns());
	}
	static double nd(){
		return Double.parseDouble(ns());
	}
	static String ns(){
		while (st == null || !st.hasMoreElements()){
			try{
				st = new StringTokenizer(br.readLine());
			}catch (IOException  e){
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}
	static int[] nia(int n){
		int[] arr=new int[n];
		for(int i=0;i<n;i++)arr[i]=ni();
		return arr;
	}
	static long[] nla(int n){
		long[] arr=new long[n];
		for(int i=0;i<n;i++)arr[i]=nl();
		return arr;
	}
	static String nline(){
		String str = "";
        try{
            str = br.readLine();
        }catch (IOException e){
            e.printStackTrace();
        }
        return str;
	}
	static ArrayList<Integer>[] nAdjList(int n,int k){
		ArrayList<Integer>[] al=new ArrayList[n];
		for(int i=0;i<n;i++)al[i]=new ArrayList<Integer>();
		for(int i=0;i<k;i++){
			int x=ni(),y=ni();
			al[x].add(y);
			al[y].add(x);
		}
		return al;
	}
	static int i(long l){
		return (int)Math.round(l);
	}
	static class Pair implements Comparable<Pair>{
		long fs,sc,val;
		Pair(long a,long b){
			fs=a;sc=b;
		}
		Pair(long a,long b,long v){
			fs=a;sc=b;val=v;
		}
		public int compareTo(Pair p){
			return i(this.val-p.val);
		}
	}
	static class MultiSet{
		public TreeSet<Pair> set;
		MultiSet(){
			set=new TreeSet<Pair>();
		}
		public void add(long x){
			if(set.contains(new Pair(1,0,x))==false)set.add(new Pair(1,0,x));
			else{
				Pair p=set.ceiling(new Pair(1,0,x));
				if(p==null || p.val!=x)set.add(new Pair(1,0,x));
				else{
					p.fs+=1;
					set.add(p);
				}
			}
		}
		public boolean contains(long x){
			return set.contains(new Pair(1,0,x));
		}
		public void print(){
			for(Pair p : set){
				out.println(p.val+" "+p.fs);
			}
		}
		public boolean remove(long x){
			if(set.contains(new Pair(1,0,x))==false)return false;
			else{
				Pair p=set.ceiling(new Pair(1,0,x));
				if(p==null || p.val!=x)return false;
				if(p.fs>1){p.fs-=1;set.add(p);return true;}
				else {set.remove(new Pair(1,0,x));return false;}
			}
		}
		public long first(){
			return set.first().val;
		}
		public long last(){
			return set.last().val;
		}
		public long ceiling(long x){
			Pair p=set.ceiling(new Pair(1,0,x));
			if(p==null)return -1L;
			else return p.val;
		}
		public long floor(long x){
			Pair p=set.floor(new Pair(1,0,x));
			if(p==null)return -1L;
			else return p.val;
		}
	}

}