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.StringTokenizer;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
	public class Main{
	
	
	public static void main(String[]args)
	{
		FastReader in = new FastReader();
		PrintWriter b=new PrintWriter(new BufferedOutputStream(System.out));
		int n=in.nextInt();
		int prev=-1,current,next;
		int prevCount=0,currentCount=1;
		int longest=1;
		current=in.nextInt();
		for(int i=1;i<n;i++)
		{
			
			next=in.nextInt();
			
			if(current==next)
				currentCount++;
			else if(next==prev)
			{
				prevCount += currentCount;
				prev = current;
				current = next;
				currentCount = 1;

				
				
			}
			else 
			{
				longest = Math.max(longest, currentCount + prevCount);
				

				prev = current;
				prevCount = currentCount;
				current = next;
				currentCount = 1;
			}
				

			
			
		}
		
		
		
	b.println(Math.max(longest, currentCount + prevCount));


		
		
		
		b.close();
	}}
class FastReader{
	BufferedReader br;
	StringTokenizer st;
	public FastReader()
	{
		br=new BufferedReader(new InputStreamReader(System.in));
		
	}
	
	String next() {
		while(st==null||!st.hasMoreElements()) {
		try {
			st=new StringTokenizer(br.readLine());
		
		} catch (IOException e) {
			e.printStackTrace();
		}
		
		}
		
		return st.nextToken();
		
		
	}
	int nextInt() {
		
		return Integer.parseInt(next()); 
	}
	double nextDouble() {
		
		return Double.parseDouble(next());
	}
	long nextLong() {
		
		return Long.parseLong(next());
	}
	String nextLine() {
		String str = "";
		try {
			str=br.readLine();
			
		} catch (IOException e) {

			e.printStackTrace();
		}
		return str;

	}

}