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;

/**
 * Created by chen on 17-10-23.
 */
public class ApproximatingConstantRange {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int n = cin.nextInt();
        int[] array = new int[n];
        for(int i = 0;i < n;i ++){
            array[i] = cin.nextInt();
        }
        int l = 0;
        int min = array[0];
        int min_index = 0;
        int max = array[0];
        int max_index = 0;
        int len = 0;
        for(int r = 0;r < n;r ++){
            if(array[r] <= min){
                min = array[r];
                min_index = r;
            }
            if(array[r] >= max){
                max = array[r];
                max_index = r;
            }
            if(max - min <= 1)
                len = Math.max(len,r - l + 1);
            else{
                if(min_index < max_index) {
                    l = min_index + 1;
                    while (max - array[l] > 1)
                        l++;
                    min = array[l];
                    min_index = l;
                }else{
                    l = max_index + 1;
                    while (array[l] - min > 1)
                        l++;
                    max = array[l];
                    max_index = l;
                }
                len = Math.max(len, r - l + 1);
            }
        }
        System.out.println(len);
    }

}
