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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigDecimal;
import java.util.*;

public class ForcesC {

	
	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		//it reads the data about the specified point and divide the data about it ,it is quite fast
		//than using direct 

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());//converts string to integer
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception e) {
				e.printStackTrace();
			}
			return str;
		}
	}


	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	static long mod = (long)(1e9+7);


	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		FastReader sc = new FastReader();
		 int n = sc.nextInt();
		
	        long[] a = new long[n];
	        for(int i=0;i<n;++i) a[i] = sc.nextLong();
	        int st = 0,ctr = 0,min = Integer.MAX_VALUE,max = Integer.MIN_VALUE;
	        long ans = 0;
	        boolean f = true;
	        TreeMap<Long,Integer> map = new TreeMap<>();
	     for(int i=0;i<n;) {
	    	 if(f)
	    	if(map.containsKey(a[i]))
	    		map.put(a[i],map.get(a[i])+1);
	    	else map.put(a[i], 1);
	    	
	    	 if(Math.abs(map.lastKey() - map.firstKey())<=1) {
	    		 ++ctr;
	    		 ++i;
	    		 f = true;
	    		 
	    	 }else {
	    		 f = false;
	    		
	    		 while(map.lastKey()-map.firstKey()>1) {
	    			// System.out.println(map.lastKey()-map.firstKey()+" : " +st);
	    			 if(map.get(a[st]) == 1)
	    				 map.remove(a[st]);
	    			 else map.replace(a[st], map.get(a[st])-1);
	    		 --ctr;
	    		 ++st;
	    		 }
	    		 
	    	 }
	    	 ans = Math.max(ans, ctr);
	     }
	      out.println(ans);
        out.close();
}
	
	static class Pair{
		int x;
		int y;
		Pair(int x,int y){
			this.x = x;
			this.y = y;
		}
	}
}