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.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Ideone
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner in=new Scanner(System.in);
		int n=in.nextInt();
		int a[]=new int[n];
		for(int i=0;i<n;i++){
			a[i]=in.nextInt();
		}
		HashMap<Integer,Integer> map=new HashMap<>();
		map.put(a[0],0);
		int len=1;
		int l=0;
		int r=1;
		int max=1;
		while(l<n && r<n){
			
			if(map.containsKey(a[r]-2) ){
			int u=map.get(a[r]-2);
		l=u+1;
		map.remove(a[r]-2);
				
			}
			if(map.containsKey(a[r]+2)){
				
				
				int u=map.get(a[r]+2);
		l=u+1;
		map.remove(a[r]+2);	
			}
			if(map.containsKey(a[r])|| map.containsKey(a[r]+1)||map.containsKey(a[r]-1)){
				
			}
			else{
				l=r;
				map=new HashMap<>();
			
			
			}
			map.put(a[r],r);
			len=(r-l)+1;
			if(len>max){
				max=len;
			}
			r++;
			
		}
		
		
		System.out.println(max);
		
	}
}