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.*;
import java.io.*;

public class Main {
    public static void main(String[] args) {
        QuickScanner sc = new QuickScanner();
        int n = sc.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextInt();
        }


        HashMap<Integer, Integer> count = new HashMap<>();
        count.put(arr[0], 1);
        int max = 0;
        int i = 0;
        int j = 1;
        while (j < n) {
            while (j < n && count.size() <= 2) {
                int v = count.containsKey(arr[j]) ? count.get(arr[j]) : 0;
                count.put(arr[j], v + 1);
                j++;
            }

            int length = j - i;

            if (count.size() > 2) {
                length--;
            }
            if (length > max) {
                max = length;
            }

            if (j < n) {
                while (i < j && count.size() > 2) {
                    int v = count.get(arr[i]) - 1;
                    if (v <= 0) {
                        count.remove(arr[i]);
                    }
                    else {
                        count.put(arr[i], v);
                    }
                    i++;
                }
            }
        }
        System.out.println(max);
    }

    private static class QuickScanner {
        BufferedReader in;
        StringTokenizer token;

        public QuickScanner() {
            this.in=new BufferedReader(new InputStreamReader(System.in));
            this.token=new StringTokenizer(""," \n\t");
        }

        public boolean hasNext() {
            while(!token.hasMoreTokens()) {
                try {
                    String s=in.readLine();
                    if(s==null) return false;
                    token=new StringTokenizer(s," \n\t");
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
            }
            return true;
        }

        public String nextLine() {
            try {
                return in.readLine();
            } catch (IOException e) {
                throw new InputMismatchException();
            }
        }

        public String next() {
            hasNext();
            return token.nextToken();
        }

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