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.util.*;
import java.io.*;

public class marte{


private int contor=0;

public static void main (String args[]) throws Exception{

Scanner input=new Scanner(System.in);

HashMap<Integer, Integer> map= new HashMap<Integer, Integer>();
HashMap<String, Integer> map2= new HashMap<String, Integer>();
HashMap<Integer, Integer> finish= new HashMap<Integer, Integer>();
HashMap<Integer, Integer> maxime= new HashMap<Integer, Integer>();
	
	int n=input.nextInt();
	int []valoare=new int[n];
	int []v=new int[n];
	
	
		for(int i=0;i<n;i++) v[i]=input.nextInt();
		
		valoare[0]=1;
		int max=v[0];
		int min=v[0];
		int pozmin=0;
		 int pozmax=0;
		 int poz=0;
		 
		 while(poz<n){
		   
		 
		 while(poz+1<n && v[poz+1]>=min && v[poz+1]<=max)
		 {
		 
		 poz++;
		 valoare[poz]=valoare[poz-1]+1;
		 if(v[poz]==min) pozmin=poz;
		 if(v[poz]==max) pozmax=poz;
		 
		 
		 
		 }
		 
		 if(poz+1<n && v[poz+1]>max && v[poz+1]-min==1){
		 
		 poz++;
		 max=v[poz];
		 
		 min=v[poz-1];
		 
		 pozmin=poz-1;
		 
		 pozmax=poz;
		 
		 valoare[poz]=valoare[poz-1]+1;
		 
		 
		 
		 
		 }
		 
		 else  if(poz+1<n && v[poz+1]<min && max-v[poz+1]==1){
		 
		 poz++;
		 min=v[poz];
		 
		 max=v[poz-1];
		 
		 pozmax=poz-1;
		 
		 pozmin=poz;
		 
		 valoare[poz]=valoare[poz-1]+1;
		 
		 
		 
		 
		 }
		 
		 else  if(poz+1<n && v[poz+1]<min && max-v[poz+1]>1 && v[poz+1]==min-1){
		 
		 poz++;
		 
		 valoare[poz]=poz-pozmax;
		 
		 min=v[poz];
		 
		 max=v[poz-1];
		 
		 pozmax=poz-1;
		 
		 pozmin=poz;
		 
		 
		 
		 
		 
		 
		 }
		 
		 else  if(poz+1<n && v[poz+1]>max && v[poz+1]-min>1 && v[poz+1]==max+1){
		 
		 poz++;
		 
		 valoare[poz]=poz-pozmin;
		 
		 min=v[poz-1];
		 
		 max=v[poz];
		 
		 pozmax=poz;
		 
		 pozmin=poz-1;
		 
		 
		 
		 
		 
		 
		 }
		 
		 else  if(poz+1<n && v[poz+1]>max+1 && max!=min){
		 
		 poz++;
		 
		 min=v[poz];
		 
		 max=v[poz];
		 
		  valoare[poz]=1;
		 
		 
		 pozmax=poz;
		 
		 pozmin=poz;
		 
		 }
		 
		 
		 else  if(poz+1<n && v[poz+1]<min-1 && min!=max){
		 
		 poz++;
		 
		 min=v[poz];
		 
		 max=v[poz];
		 
		  valoare[poz]=1;
		 
		 
		 pozmax=poz;
		 
		 pozmin=poz;
		 
		 }
		 
		 
		 
		   if(poz==n-1) poz=n;
		
		 
		 
		 
		 
		 
		 }
		 int pozcontor=0;
		 int contor=0;
		 
		 for(int i=0;i<n;i++){ contor=Math.max(contor,valoare[i]);
		 pozcontor=i;
		 }
		 
		 
		

		 
	
	System.out.println(contor);
	
	
		}
	
}
	

	
	
	
	
	
	