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.InputStreamReader;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Dictionary;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Hashtable;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

import javax.print.attribute.DocAttributeSet;
import javax.print.attribute.HashDocAttributeSet;

import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
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);
		TaskA solver = new TaskA();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskA {
	 
	 public void solve(int testNumber, InputReader in, PrintWriter out) {
		 int n = in.nextInt();
		 int[] a = in.nextIntegerArray(n);
		 int[][] dp = new int[2][n];
		 dp[0][0] = dp[1][0] = 1;
		 for(int i = 1; i < n; i++){
			 if(a[i] == a[i-1]){
				 dp[0][i] = 1 + dp[0][i-1];
				 dp[1][i] = 1 + dp[1][i-1];
			 }else if(a[i] < a[i-1]){
				 dp[0][i] = 1 + dp[1][i-1];
				 dp[1][i] = 1;
			 }else{
				 dp[1][i] = 1 + dp[0][i-1];
				 dp[0][i] = 1;
			 }
		 }
		 int max = -(int) (1/0.0);
		 for(int i = 0; i < n; i++){
			 //out.println(dp[0][i] + " " + dp[1][i]);
			 max = Math.max(max, Math.max(dp[0][i], dp[1][i]));
		 }
		 out.println(max);
	 }
}
  
  
         //||||||| comparable pair ||||||||
 class pair implements Comparable<pair>{
	 public long first,second;
	 public pair(){
		 first = second = 0;
	 }
	@Override
	public int compareTo(pair arg0) {
		if(this.first > arg0.first)
			return 1;
		else if(this.first < arg0.first)
			return -1;
		else if(this.second > arg0.second)
			return 1;
		else if(this.second < arg0.second)
		     return -1;
		return 0;
	}
 }

 
   // ||||||| INPUT READER ||||||||
class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream));
        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 Integer nextInt() {
        return Integer.parseInt(next());
    }
    
    public Long nextLong() {
    	return Long.parseLong(next());
    }
    
    public Double nextDouble() {
    	return Double.parseDouble(next());
    }
    
    public Float nextFloat() {
    	return Float.parseFloat(next());
    }
    
    public BigInteger nextBigInteger() {
    	return new BigInteger(next());
    }
    
    public String nextLine() {
    	try {
			return reader.readLine();
		} catch (IOException e) {
			// TODO Auto-generated catch block
			e.printStackTrace();
			return null;
		}
    	
    }
    
    public int[] nextIntegerArray(int n) {
    	int[] a = new int[n];
    	for(int i = 0; i <  n; i++)
    		a[i] = nextInt();
    	
    	return a;
    }
    
    public Long[] nextLongArray(int n) {
    	Long[] a = new Long[n];
    	for(int i = 0; i <  n; i++)
    		a[i] = nextLong();
    	
    	return a;
    }
    
    public long[] nextlongArray(int n) {
    	long[] a = new long[n];
    	for(int i = 0; i <  n; i++)
    		a[i] = nextLong();
    	
    	return a;
    }
    
    public double[] nextdoubleArray(int n) {
    	double[] a = new double[n];
    	for(int i = 0; i <  n; i++)
    		a[i] = nextDouble();
    	
    	return a;
    }
    
    public String[] nextStringArray(int n){
    	String[] s = new String[n];
    	for(int i = 0; i < n; i++)
    		s[i] = next();
    	return s;
    }
    
    public String[] nextLineStringArray(int n){
    	String[] s = new String[n];
    	for(int i = 0; i < n; i++)
    		s[i] = next();
    	return s;
    }

    
    }