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

/**
 * Created by yujiahao on 3/20/16.
 */
public class cf_333_b {
    private FastScanner in;
    private PrintWriter out;

    public void solve() throws IOException {
        int n = in.nextInt();
        int[] a = new int[n];
        for (int i=0; i<n; i++){
            a[i] = in.nextInt();
        }

        int prev_diff = 0;
        int cur_diff  = 0;
        int l = 0;
        int r = 0;
        int res = 0;
        boolean allSame = true;

        for (int i=1; i<n; i++){
            cur_diff = a[i] - a[i-1];

            if (cur_diff==0){
                continue;
            }

            allSame = false;

            if (cur_diff!=prev_diff){
                r = i;
                prev_diff = cur_diff;
            }else{ // the same difference: -1 -1 or +1 +1
                res = Math.max(res, i-l);
                l = r;
                r = i;
            }
        }

        if (allSame){
            out.print(n);
        }else{
            out.print(Math.max(res, n-l));
        }
    }


    public void run() {
        try {
            in = new FastScanner();
            out = new PrintWriter(System.out);
            solve();
            out.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    private class FastScanner {
        private BufferedReader br;
        private StringTokenizer st;

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

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

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

        public long nextLong() {
            return Long.parseLong(next());
        }

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

        public BigInteger nextBigInteger() { return new BigInteger(next());}

    }

    public static void main(String[] arg) {
        new cf_333_b().run();
    }
}
