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.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String args[]){
        int n;
        int[] dp=new int[100015];
        int[] a=new int[100015];
        Scanner in=new Scanner(System.in);
        while(in.hasNext()){
            n=in.nextInt();
            for(int i=1;i<=n;i++)
            {
                a[i]=in.nextInt();
            }
            Arrays.fill(dp,0);
            dp[a[1]]=1;
            int hehe=1;
            int ans=-1;
            for(int i=2;i<=n;i++)
            {
                if(a[i]>2&&dp[a[i]-2]==0&&dp[a[i]+2]==0||a[i]<=2&&dp[a[i]+2]==0){
                    hehe=Math.min(hehe+1,i);
                }
                else{
                    int index=-1;
                    if(a[i]>2)
                        index=Math.max(dp[a[i]-2],dp[a[i]+2]);
                    else
                        index=dp[a[i]+2];
                    hehe=Math.min(hehe+1,i-index);
                }
                ans=Math.max(ans,hehe);
                dp[a[i]]=i;
            }
            System.out.println(ans);
        }
    }
}
