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.*;

 public class Dcoder
 {
	public static void main(String args[])
 	{
 		Scanner input = new Scanner(System.in);
 		
 		int n = input.nextInt() ;
 		int[] arr = new int[n] ;
 		int[] brr = new int[100001];
 		
 		for(int i = 0 ; i < n ; i++)
 		{
 			arr[i] = input.nextInt();
 		}
 		
 		int i,j;
 		i = j = 0 ;
 		
 		int len = 1 ;
 		int min,max ;
 		min = max = arr[0] ;
 		
 		for(;j < n ; j++)
 		{
 			brr[arr[j]]++ ;
 			
 			if(arr[j] < min)
 			{
 				min = arr[j] ;
 			}
 			if(arr[j] > max)
 			{
 				max = arr[j] ;
 			}
 			
 			for(;i < j && max - min > 1; i++)
 			{
 				if(max == arr[j] && brr[min] == 0)
 				{
 					for(int k = min ;brr[min] == 0;k++)
 					{
 						min = k ;
 					}
 				}
 				if(min == arr[j] && brr[max] == 0)
 				{
 					for(int k = max ;brr[max] == 0 ; k--)
 					{
 						max = k ;
 					}
 				}
 				
 				brr[arr[i]]-- ;
 				if(max - min <= 1)
 				{
 					brr[arr[i]]++ ;
 					i-- ;
 				}
 			}
 			if(j-i+1 > len )
 			{
 				len = j-i+1 ;
 			}
 		}
 		System.out.print(len);
 	}
 }
