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.Scanner;
public class GYM {
    private static Scanner in = new Scanner (System.in);
    private static int N,C,ans,xp[],val,fun,max=Integer.MIN_VALUE,min=Integer.MAX_VALUE;
    public static void main(String[] args) {
        N=in.nextInt();xp=new int [N];
       for(int i =0;i<N;i++){xp[i]=in.nextInt();}
       for(int i =0;i<N;i++){if(i!=0&&(xp[i]==min||xp[i]==max)){}else{max=Integer.MIN_VALUE;min=Integer.MAX_VALUE;C=0; sum(i);} if(C>ans)ans=C;}
        System.out.println(ans);
       }
private static int sum(int o)
{
    if(o<N){if(max<xp[o]){max=xp[o];}
if(min>xp[o]){min=xp[o];}
if(max-min<=1)
{C++;
for(int i =o+1;i<N;i++)
{if(min==xp[i]||max==xp[i]){C++;}
else {sum(i);break;}
}
 }}
    { return C;}}}