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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

import static java.lang.Math.max;
import static java.lang.Math.min;

public class Code_602B_ApproximatingAConstantRange {
    public static void main(String[] args) throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out=new PrintWriter(System.out);
        StringTokenizer in=new StringTokenizer(br.readLine());
        int n=Integer.parseInt(in.nextToken());
        int[] a=new int[n];
        in=new StringTokenizer(br.readLine());
        for(int i=0;i<n;a[i++]=Integer.parseInt(in.nextToken()));
        int l=0,r=0;
        int length=1,largest=a[l],smallest=a[r];
        while (l<=r && r<n){
            largest=max(largest,a[r]);
            smallest=min(smallest,a[r]);
            if(largest-smallest<=1) {
                length = max(length, r - l + 1);
                ++r;
            }else{
                l=r-1;
                int value=a[l];
                while (l>0 && a[l-1]==value)--l;
                smallest=min(a[l],a[r]);
                largest=max(a[l],a[r]);
            }
        }

        out.println(length);
        out.close();
    }
}
