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 Hue {

    static class Solution { 
        InputReader in;
        PrintWriter out;

        public void solve() {
            int n = in.nextInt();
            int ans = 0;
            int map[] = new int[100111];

            for (int i=0; i<map.length; i++) map[i] = -1;

            for (int i=0; i<n; i++) {
                int idx  = n;
                int num  = in.nextInt();
                map[num] = i;

                if (map[num+1] > map[num-1]) ans = Math.max(ans,i-Math.max(map[num+2],map[num-1])); else
                if (map[num+1] < map[num-1]) ans = Math.max(ans,i-Math.max(map[num-2],map[num+1])); else
                ans = Math.max(ans,i+1);                
            }

            out.println(ans);
        }
    }

    public static void main(String args[]) {
        InputReader in    = new InputReader(System.in);
        PrintWriter out   = new PrintWriter(System.out);
        Solution solution = new Solution();
        solution.in       = in;
        solution.out      = out;
        solution.solve();
        out.close();
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

        public long nextLong() { return Long.parseLong(next()); }
    
        public String nextLine() {
            String ret = "";
            try { ret = reader.readLine(); } catch(Exception e) {}
            return ret;
        }
    }
}