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.FileNotFoundException;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;

/**
 * Created by slycer on 10/15/15.
 */
public class P2 {

    public static void main ( String [] args ) throws IOException {
        BufferedReader reader = new BufferedReader( new InputStreamReader( System.in ) );
        int n = Integer.valueOf( reader.readLine() );
        String [] aux = reader.readLine().split( " " );
        int [] data = new int[n];
        for ( int i=0; i<n; i++ ){
            data[i] = Integer.valueOf( aux[i] );
        }
        int left = 1;
        int right = data.length;
        while ( left <= right ){
            int m = ( left + right )/2;
            boolean ok = rangeAlmostConstant( data, m );
            //System.out.println( ok + " " + m );
            if ( ok ){
                left=m+1;
            }
            if ( !ok ){
                right=m-1;
            }
        }

        System.out.println( right );

    }

    public static boolean rangeAlmostConstant( int [] data, int rangeLength ){
        MinMaxStructure minmax = new MinMaxStructure();
        for ( int i=0; i<data.length; i++ ){
            minmax.add( data[i] );
            if ( i >= rangeLength ){
                minmax.remove( data[i-rangeLength] );
            }
            if ( i >= rangeLength-1 ){
                if ( minmax.max - minmax.min <= 1 ){
                    return true;
                }
            }
           // System.out.println( data[i] + " " + rangeLength + " " + minmax.max + " " + minmax.min );
        }
       // System.out.println();
        return false;
    }


    public static class MinMaxStructure {
        private int [] data = new int[100001];
        int count = 0;
        int min = -1;
        int max = -1;

        public MinMaxStructure( ){
        }

        public void add( int value ){
            data[value]++;
            if ( min == -1 || value < min ){
                min = value;
            }
            if ( max == -1 || value > max ){
                max = value;
            }
        }

        public void remove( int value ){
            data[value]--;
            if ( data[value] == 0 ) {
                if (value == min) {
                    min++;
                }
                if (value == max) {
                    max--;
                }
            }
        }

    }


}
