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].





/* package codechef; // don't place package name! */
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.StringTokenizer; 
import java.util.*;

public class HelloWorld{

    public static void main(String[] args) throws IOException 
    { 
  
        Scanner input = new Scanner(System.in);
        while(input.hasNext()){
            int n=input.nextInt();
            int[] l=new int[n];
            int[][] dp=new int[n][2];
            for(int i=0;i<n;i++){
                l[i]=input.nextInt();
            }
            dp[n-1][0]=1;
            dp[n-1][1]=1;
            int max=1;
            for(int i=dp.length-2;i>=0;i--){
                if(l[i]==l[i+1]){
                    dp[i][0]=dp[i+1][0]+1;
                    dp[i][1]=dp[i+1][1]+1;
                    max=Math.max(max, Math.max(dp[i][0], dp[i][1]));
                    continue;
                }
                dp[i][0]=1;
                if(i+1+dp[i+1][0]>n-1){
                    dp[i][1]=1+dp[i+1][1];
                    max=Math.max(max, Math.max(dp[i][0], dp[i][1]));
                    continue;
                }
                if(l[i+1+dp[i+1][0]]==l[i]){
                    dp[i][1]=1+dp[i+1][1];
                    max=Math.max(max, Math.max(dp[i][0], dp[i][1]));
                    continue;
                }
                dp[i][1]=1+dp[i+1][0];
                max=Math.max(max, Math.max(dp[i][0], dp[i][1]));
            }
            System.out.println(max);
        }
    }

}