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) {
		try (Scanner scan = new Scanner(System.in)) {
			
			int N = scan.nextInt();
			int[] list = new int[N];
			for (int i=0;i<N;i++)list[i]=scan.nextInt();
			int maxsize = 1;
			int min=list[0], max = list[0],  size = 1;
			int idxOfMin=0, idxOfMax=0;
			for (int i=1;i<list.length;i++){
				if (list[i]>max)max=list[i];
				if (list[i]<=min){min=list[i];}
				if (Math.abs(min-list[i])<=1 && Math.abs(max-list[i])<=1){
					size++;
					if (list[i]==min)idxOfMin=i;
					if (list[i]==max)idxOfMax=i;
					maxsize=Math.max(size, maxsize);
				}else if (list[i] == max){
					size=1;
					i=idxOfMin+1;
					if (i==list.length)break;
					min=list[i];max=list[i];
					idxOfMin=i;
					idxOfMax=i;
				}else {
					size=1;
					i=idxOfMax+1;
					if (i==list.length)break;
					min = list[i];max=list[i];
					idxOfMin=i;
					idxOfMax=i;
				}
			}
			System.out.println(maxsize);
		}
	}
}