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 Bai2_Extra_602B_ApproAConstantRange {
    /*https://codeforces.com/problemset/problem/602/B*/
    public static void main(String[] args) {
        /*
        ông đã từng phải đo cường độ của một hiệu ứng từ từ tiến đến trạng thái cân bằng
        choosing a sufficiently large number of consecutive-lien tiep data points
         */

        Scanner sc = new Scanner(System.in);
        int n; // number of sequence data
        n = sc.nextInt();

        int[] arrDataPoint = new int[n];
        for (int i = 0; i < n; i++) {
            arrDataPoint[i] = sc.nextInt();
        }
        //data points not have big jum : |a_(i + 1) - a_i| <= 1
        // rang [l,r] -> almost constant if biggest(M) - smallest(m) <= 1

        //output: maximum length of an almost constant
        int maxLenght = 0;
        int count = 0;
        int max = arrDataPoint[0];
        int min = arrDataPoint[0];
        for (int i = 0; i < n; i++) {
            // voi k = i, constant amount co the cong don tiep tuc tu i tro di
            if (Math.abs(arrDataPoint[i] - min) <= 1 && Math.abs(arrDataPoint[i] - max) <= 1) { // tim duoc doan constant tu vi tri i
                count++;
                min = Math.min(arrDataPoint[i], min);
                max = Math.max(arrDataPoint[i], max);
            } else {
                maxLenght = Math.max(maxLenght, count);
                //cap nhat dieu kien min max tai vi tri i
                min = arrDataPoint[i];
                max = arrDataPoint[i];
                count = 0;
                // tinh so luong constant tu vi tri thu i ve truoc do (co kha nang cong don voi i tiep theo)
                for (int k = i; k >= 0; k--) {
                    if (Math.abs(arrDataPoint[k] - min) <= 1 && Math.abs(arrDataPoint[k] - max) <= 1) {
                        count++;
                        min = Math.min(arrDataPoint[k], min);
                        max = Math.max(arrDataPoint[k], max);
                    }
                    else break;
                }
            }
        }
        System.out.println(Math.max(maxLenght,count));
    }// 1 2 3 4
}
