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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
import java.util.List;
import java.util.ArrayList;
import java.util.Collections;


public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(in, out);
        out.close();
    }
    
    static class TaskB {
    	 public void solve(InputReader in, PrintWriter out) {
    		 int n = in.nextInt();
    		 int next;
    		 int last;
    		 int streak = 1;
    		 int maxLen = 2;
    		 int len = 1;
    		 boolean lower = false;
    		 boolean higher = false;
    		 last = in.nextInt();
    		 for (int i = 1; i < n; i++) {
    			 next = in.nextInt();
    			 if (next == last) {
    				 streak += 1;
    				 len += 1;
    			 }
    			 else if (next > last) {
    				 last = next;
    				 if (!higher) {
    					 len += 1;
    					 higher = true;
    					 lower = false;
    				 }
    				 else if (higher) {
    					 len = streak + 1;
    				 }
    				 streak = 1;
    			 }
    			 else {
    				 last = next;
    				 if (!lower) {
    					 len += 1;
    					 lower = true;
    					 higher = false;
    				 }
    				 else if (lower) {
    					 len = streak + 1;
    				 }
    				 streak = 1;
    			 }
    			 if (len > maxLen)
    				 maxLen = len;
    		 } 
    		 out.println(maxLen);
        }	   	
    }       
    
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        
        public long nextLong() {
        	return Long.parseLong(next());
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
    }
}