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.math.BigInteger;
import java.util.*;

import javafx.util.Pair;

public class Main
{

    public static void main(String[] args)
    {
        FastScanner input = new FastScanner();

        int n = input.nextInt();
        int a[] = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = input.nextInt();

        }
        int ans = 1;
        int min = Integer.MAX_VALUE;
        int max = Integer.MIN_VALUE;
        int nextMinIndex = 0;
        int nextMaxIndex=0;
        int count = 0;
        for (int i = 0; i < n; i++) {

            if (a[i] <= min) {
                nextMinIndex = i;
                min = a[i];
            }
            if(a[i]>=max)
            {
                nextMaxIndex=i;
                max = Math.max(max, a[i]);
            }
            
            if (Math.abs(max - min) <= 1) {
                 count++;
                ans = Math.max(ans, count);
               
            } else {
                count = 0;
                min = Integer.MAX_VALUE;
                max = Integer.MIN_VALUE;
                i = Math.min(nextMaxIndex, nextMinIndex);
            }
        }
        System.out.println(ans);

    }

    static class FastScanner
    {

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

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

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {

            return Long.parseLong(next());
        }

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

        String nextLine() throws IOException
        {
            return br.readLine();
        }
    }

}
