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.InputStreamReader;
import java.util.StringTokenizer;

public class R333Div2B {
	public static void main(String[] args) {
		FastScanner in=new FastScanner();
		int n=in.nextInt();
		int[] a=new int[n];
		for(int i=0;i<n;i++)
			a[i]=in.nextInt();
		
		int[] dif=new int[n-1];
		for(int i=0;i<n-1;i++)
			dif[i]=a[i+1]-a[i];
		
		int last=0;
		int max=0;
		int idx=-1;
		int i,j;
		for(i=0,j=0;i<n-1;i++){
			if(dif[i]==1&&last!=1){
				last=1;
				idx=i;
			}else if(dif[i]==-1&&last!=-1){
				last=-1;
				idx=i;
			}else if(dif[i]==0){
			}else{
				max=Math.max(max,i-j+1);
				j=idx+1;
				if(dif[i]==1){
					last=1;
					idx=i;
				}else if(dif[i]==-1){
					last=-1;
					idx=i;
				}else{
					last=0;
					idx=-1;
				}
			}
		}
		if(j<i){
			max=Math.max(max,i-j+1);
		}
		System.out.println(max);
	}
	static class FastScanner{
		BufferedReader br;
		StringTokenizer st;
		public FastScanner(){br=new BufferedReader(new InputStreamReader(System.in));}
		String nextToken(){
			while(st==null||!st.hasMoreElements())
				try{st=new StringTokenizer(br.readLine());}catch(Exception e){}
			return st.nextToken();
		}
		int nextInt(){return Integer.parseInt(nextToken());}
		long nextLong(){return Long.parseLong(nextToken());}
		double nextDouble(){return Double.parseDouble(nextToken());}
	}
}
