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.io.BufferedInputStream;
import java.util.Scanner;

/**
 * Created by ADMIN on 4/27/2017.
 */
public class ApproximatingAConstantRange {
    static Scanner sc = new Scanner(new BufferedInputStream(System.in));

    public static void main(String[] args) {
        int a[] = new int[100000];
        int n = sc.nextInt();
        for (int i = 0; i < n; i++) a[i] = sc.nextInt();
        int minIdx = 0, maxIdx = 0, min = a[0], max = a[0], i = 0, j = 0, best = 1;
        while (j < n - 1) {
            j++;
            if(max == min)
            {
                if(a[j]<min)
                {
                    min = a[j];
                    minIdx = j;
                }
                else if(a[j]>max){
                    max = a[j];
                    maxIdx = j;
                }
                else
                {
                    max = min = a[j];
                    minIdx = maxIdx = j;
                }
            }
            else if (a[j] == max)
                maxIdx = j;
            else if (a[j] == min)
                minIdx = j;
            else if (a[j] - max == 1 && a[j-1]==max) {
                i = minIdx+1;
                min = max;
                max = a[j];
                maxIdx = j;
                minIdx = j-1;
            } else if (min - a[j] == 1 && a[j-1]==min) {
                i = maxIdx+1;
                max = min;
                min = a[j];
                minIdx = j;
                maxIdx = j-1;
            }
            else
            {
                min = max = a[j];
                maxIdx = minIdx = i = j;
            }

            //System.out.printf("current range is (%d,%d)\n",i,j);

            best = Math.max(best, j - i + 1);
        }
        System.out.println(best);
    }

}
