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.util.StringTokenizer;


public class ConstantRange {
    static int n;
    static int a[];
    public static void main(String[] args) throws NumberFormatException, IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        n = Integer.parseInt(br.readLine());
        StringTokenizer st = new StringTokenizer(br.readLine());
        a = new int[n];
        for(int i=0; i<n; i++)
            a[i] = Integer.parseInt(st.nextToken());

        int p1 = 0;

        int num1 = a[0];
        int num2 = a[1];
        int max = 2;
        int tmpMax = 2;
        for(int i=2; i<n; i++){
            if(a[i] != num2 && a[i] != num1){
                if(num1 == num2)
                {
                    
                        num2 = a[i];
                        tmpMax++;
                        max = Math.max(max,tmpMax);
                        
                    
                    continue;
                }
                int j = i-1;
                num2 = a[i];
                int t = num1;
                while(Math.abs(a[i] - a[j])<=1){
                    t = a[j];
                    j--;
                }
                num1 = t;
                j++;
                p1 = j;

                tmpMax = i-p1+1;
                //System.out.println(p1+" "+i+" "+num2+" "+num1);
            }
            else{
                tmpMax++;
            }
            max = Math.max(tmpMax,max);
        }
        System.out.println(max);



    }


}
