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 B {

    public static void main(String[] args) 
    {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int [] X = new int [n];
        for(int i=0;i<n;i++)
        {
            X[i] = sc.nextInt();
        }
        int leng = 1;
        int max_leng = leng;
        int max = X[0];
        int min = X[0];
        int l = 0;
        int r = 0;
        int start = 0;
        //int end = 0;
        for(int i=1; i<n; i++)
        {
            if(Math.abs(min-X[i])<=1 && Math.abs(max-X[i])<=1)
            {
                leng++;
                if (X[i]>=max)
                {
                    max = X[i];
                    r = i;
                }
                else if (X[i]<=min)
                {
                    min = X[i];
                    l = i;
                }
                continue;
            }
            else
            {
                if(leng>max_leng)
                    max_leng = leng;
                
                if(X[i]>max)
                {
                    max = X[i];
                    r=i;
                    leng = leng - (l - start);
                    start = l + 1;
                    min = X[i-1];
                    l = i-1;
                }
                
                else if(X[i]<min)
                {
                    min = X[i];
                    l=i;
                    leng = leng - (r - start);
                    start = r + 1;
                    max = X[i-1];
                    r = i-1;
                }
            }
        }
        if(leng>max_leng)
            max_leng = leng;
        
        System.out.print(max_leng);
    }
    
}