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.*;
import java.util.*;
public class B602 {
public static void main(String[] args) throws IOException {
    input.init(System.in);
    PrintWriter out = new PrintWriter(System.out);
    int n = input.nextInt();
    int[] as = new int[n];
    for(int i = 0; i<n; i++) as[i] = input.nextInt();
    int[] bs = new int[n], cs = new int[n];
    for(int i = 0; i<n; i++)
    {
        bs[i] = cs[i] = as[i];
        if(as[i]%2 == 0) bs[i]++;
        else cs[i]++;
    }
    int max = 1, curb = 1, curc = 1;
    for(int i = 1; i<n; i++)
    {
        if(bs[i] == bs[i-1]) curb++;
        else curb = 1;
        if(cs[i] == cs[i-1]) curc++;
        else curc = 1;
        max = Math.max(max,  curb);
        max = Math.max(max,  curc);
    }
    out.println(max);
    out.close();
}

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

    static void init(InputStream input) {
    reader = new BufferedReader(new InputStreamReader(input));
    tokenizer = new StringTokenizer("");
    }

    static String next() throws IOException {
    while (!tokenizer.hasMoreTokens())
        tokenizer = new StringTokenizer(reader.readLine());
    return tokenizer.nextToken();
    }

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

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

    static long nextLong() throws IOException {
    return Long.parseLong(next());
    }
}
}
