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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Problem602B
{
	public static void main(String[] args) throws IOException
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		
		int[] a = new int[n];
		st = new StringTokenizer(br.readLine());
		for(int i=0 ; i<n ; i++)
		{
			a[i] =  Integer.parseInt(st.nextToken());
		}
		
		int i, maxlength=1, min=a[0], max=a[0], first=0, last=0, index2=0;
		
		for(i=1 ; i<n ; i++)
		{
			if(a[i] >= min && a[i] <= max)
			{
				if(a[i] == max)
					last = i;
				else if(a[i] == min)
					first = i;
			}
			else if(a[i] > max)
			{
				if(a[i] > max+1)
				{
					index2 = i;
					min = max = a[i];
					first = last = i;
				}
				else if(min == max)
				{
					max = a[i];
					last = i;
				}
				else
				{
					index2 = first + 1;
					min = max;
					max = a[i];
					first = last;
					last = i;
				}
			}
			else if(a[i] < min)
			{
				if(a[i] < min-1)
				{
					index2 = i;
					min = max = a[i];
					first = last = i;
				}
				else if(min == max)
				{
					min = a[i];
					first = i;
				}
				else
				{
					index2 = last + 1;
					max = min;
					min = a[i];
					last = first;
					first = i;
				}
			}
			if(maxlength < i-index2+1)
				maxlength = i-index2+1;
		}
		System.out.println(maxlength);
	}
}
