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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.StringTokenizer;

/**
*
* Codeforces Round #333 (Div 2) - B. Approximating a Constant Range
*
* In-progress
*
* @author Daniel
*
*/
public class B {
    
    static void solve() throws Exception {
        
        int n = nextInt();
        int[] a = readIntArray(n);
        
        int min = a[0];
        int max = a[0];
        int minIndex = 0;
        int maxIndex = 0;
        int longestRange = 1;
        int currentLength = 1;
        
        for (int i=1; i<n; i++) {
            int num = a[i];
            //Maintain everything is fine
            if (num == min || num == max)
                currentLength++;
            if (num == min)
                minIndex = i;
            if (num == max)
                maxIndex = i;
            //New max
            else if (num > max) {
                //Range is too long, re-adjust everything
                if (num - min > 1) {
                    currentLength = i - minIndex;
                    minIndex = maxIndex;
                    min = max;
                }
                else {
                    currentLength++;
                }
                max = num;
                maxIndex = i;
            }
            //New min
            else if (num < min) {
                if (max - num > 1) {
                    currentLength = i - maxIndex;
                    maxIndex = minIndex;
                    max = min;
                }
                else {
                    currentLength++;
                }
                min = num;
                minIndex = i;
            }
            longestRange = Math.max(longestRange, currentLength);
        }
        printf("%d\n", longestRange);
    }


    /*=============================================================================
     *                          STANDARD I/O TEMPLATE
     * ============================================================================
     */
    static BufferedReader in;
    static BufferedWriter out;
    static StringTokenizer tok;
    
    public static void main(String[] args) {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new BufferedWriter(new OutputStreamWriter(System.out));
            solve();
            out.flush();
            out.close();
            in.close();
        } catch (Throwable e) {
            e.printStackTrace();
            System.exit(1);
        }
    }
    
    static void print(CharSequence str) throws IOException {
        out.write(str.toString());
    }
    
    static void println() throws IOException {
        out.write(System.lineSeparator());
    }
    
    static void println(CharSequence str) throws IOException {
        print(str);
        println();
    }
    
    static void printf(String format, Object... args) throws IOException {
        print(String.format(format, args));
    }
    
    static int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

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

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

    static String next() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }
    
    static String nextLine() throws IOException {
        if (tok == null || !tok.hasMoreTokens())
            return in.readLine();
        else
            return tok.nextToken("\n\r");
    }
    
    static int[] readIntArray(int n) throws IOException {
        int[] a = new int[n];
        for (int i=0; i<n; i++)
            a[i] = nextInt();
        return a;
    }
}
