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;



public class Equillibrium {
    public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n=Integer.parseInt(br.readLine());
        String []s=br.readLine().split(" ");
        int []a=new int [n];

        
        for(int i=0;i<n;i++){
            a[i]=Integer.parseInt(s[i]);
        }
        int ans=0;
        int temp=1;
        int max1=a[0],min1=a[0];
        for(int j=1;j<n;j++){
            max1=Math.max(max1,a[j]);
            min1=Math.min(min1,a[j]);
            if(Math.abs(max1-min1)<=1){
                temp++;
            }
            else{
                max1=Math.max(a[j-1],a[j]);
                min1=Math.min(a[j-1],a[j]);
                if(Math.abs(max1-min1)<=1){
                   int k=j-1;
                    temp=1;
                    while(k>=0 && (a[k]==min1 || a[k]==max1)){
                        temp++;
                        k--;
                    }
                }
                else{
                    max1=a[j];
                    min1=a[j];
                    temp=1;
                }
            }
            ans=temp>ans?temp:ans;
        }
        System.out.println(ans);
    }
}
