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 Solution{
     
     public static void main(String []args){
            Scanner sc = new Scanner(System.in);
            
            int n = sc.nextInt();
           
            int[] a = new int[n + 1];
            
            for(int i = 1 ; i <= n ; i++){
                a[i] = sc.nextInt();
            }
            
            int res = 0;
            int prev = 0;
            int l = 1;
            int j = 1;
            
            
            for(int i = 2 ; i <= n ; i++){
                
                if(a[i] - a[i - 1] == 0){
                    continue;
                }
                
                if(a[i] - a[i - 1] == prev){
                    
                    res = Math.max(res , i - j);
                    j = l;
                }
                else{
                    prev = a[i] - a[i - 1];
                }
                l = i;
                
            }
           res = Math.max(res , n + 1 - j);
            System.out.println(res);
            
           
           
      
     }
     
     
}

 