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.*;
import java.lang.*;
public class Codeforces{
    BufferedReader br =new BufferedReader(new InputStreamReader(System.in)); 
    PrintWriter out =new PrintWriter(System.out);
    StringTokenizer st =new StringTokenizer("");
    String next(){
        if(!st.hasMoreTokens()){
            try{
                st=new StringTokenizer(br.readLine());
            }
            catch(Exception e){

            }
        }
        return st.nextToken();
    }
    int nextInt(){
        return Integer.parseInt(next());
    }
    public static void main(String[] args) {
        new Codeforces().solve();
    }
    void solve(){
        int n=nextInt();
        int arr[]=new int[n];
        for(int i=0;i<n;i++) arr[i]=nextInt();
        int i=0,j=0,left=0,res=0,l=0;
        for(i=1;i<n;i++){
            if(arr[i]-arr[i-1]==0){
                continue;
            }
            if(arr[i]-arr[i-1]==left){
                res=Math.max(res,i-j);
                j=l;
            }
            else{
                left=arr[i]-arr[i-1];
            }
            l=i;
        }
        out.println(Math.max(res,i-j));
        out.close();
    }
}
