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.Scanner;


public class Main {

    public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int n = input.nextInt();
int [] a= new int[n];
for (int i = 0; i < a.length; i++) {
    a[i]=input.nextInt();
    
}
int max = a[0];
int min = 1000000;
int count = 0;
int maxx=0;
for(int i = 0;i<a.length;i++){
    if(a[i]>max){
        min = max;
        max = a[i];
        break;
    }else if(a[i]<max){
        min=a[i];
        break;
    }
    
}
for(int i = 0;i<a.length;i++){
    if(a[i]<=max&&a[i]>=min){
        count++;
        if(count>maxx)maxx=count;
    }else if(a[i]>max){
        max=a[i];
        min=a[i]-1;
        count = 1;
        for(int j = i-1;j>=0;j--){
            if(a[j]<=max&&a[j]>=min){
                count++;
                if(count>=maxx)maxx=count;
            }else break;
        }
        
    }else if(a[i]<min){
        max=a[i]+1;
        min=a[i];
        count = 1;
        for(int j = i-1;j>=0;j--){
            if(a[j]<=max&&a[j]>=min){
                count++;
                if(count>maxx)maxx=count;
            }else break;
        }
        
    }
    
}

System.out.println(maxx);

    }

}
