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].
/**
 * Created by shambala on 24/11/15.
 */
import java.util.*;
import java.io.*;

public class TaskB {

    FastScanner in;
    PrintWriter out;

    public void solve() throws IOException {
        int n = in.nextInt();
        int[] mas = new int[n];
        for (int i  = 0; i<n; i++) {
            mas[i] = in.nextInt();
        }
        int[] d= new int[n-1];
        for (int i = 0; i<n-1; i++) {
            d[i] = mas[i+1]-mas[i];
        }
        int left = 0;
        int right = 0;
        int min = mas[0];
        int max = mas[0];
        int maxLen = 0;
        while (right<n-1) {
            while (right<n-1 && Math.abs(max-min)<=1) {
                right++;
                if (mas[right] > max) {
                    max = mas[right];
                }
                if (mas[right] < min) {
                    min = mas[right];
                }

            }
            if (right==n-1 && Math.abs(max-min)<=1) {
                if (right - left+1 > maxLen) {
                    maxLen = right - left+1;
                }
            }
            if (right - left > maxLen) {
                maxLen = right - left;
            }
            int val = mas[right];
            int newLeft = left;
            for (int i = left; i<right; i++) {
                if (Math.abs(mas[i] - val) > 1) {
                    newLeft = i+1;
                }
            }
            left = newLeft;
            max = mas[left];
            min = mas[left];
            for (int i = left; i<=right; i++) {
                if (mas[i]>max) {
                    max = mas[i];
                }
                if (mas[i]<min) {
                    min = mas[i];
                }
            }
        }
        System.out.println(maxLen);
    }

    public void run() {
        try {
            in = new FastScanner();
            //out = new PrintWriter(new File(".out"));

            solve();

        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(File f) {
            try {
                br = new BufferedReader(new FileReader(f));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }
        }

        FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

    public static void main(String[] arg) {
        new TaskB().run();
    }
}

