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 B { 
    public static void main(String [] args) throws IOException { 

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int N = Integer.parseInt(br.readLine());
        
        int [] arr = new int[N];


        StringTokenizer tokenizer = new StringTokenizer(br.readLine());
        
        for(int i=0; i<N; i++) { 
            int num = Integer.parseInt(tokenizer.nextToken());
            arr[i] = num;
        }

        int l = 0; 
        int r = 0;
        int maxCurrElem = arr[0];
        int minCurrElem = arr[0];
        int maxDist = 0;

        int ind = 1;

        while(ind < N) {
            if(Math.abs(maxCurrElem - arr[ind]) > 1 || Math.abs(minCurrElem - arr[ind]) > 1) { 
                maxDist = Math.max(maxDist, ind - l);
                if(Math.abs(arr[r] - arr[ind]) <= 1)  {
                    l = r;
                    r = ind;
                    maxCurrElem = Math.max(arr[l], arr[r]);
                    minCurrElem = Math.min(arr[l], arr[r]);
                } else { 
                    l = r = ind;
                    maxCurrElem = minCurrElem = arr[ind];
                }

                ind++;
                continue;
            }
            maxCurrElem = Math.max(maxCurrElem, arr[ind]);
            minCurrElem = Math.min(minCurrElem, arr[ind]);
            if(arr[ind] != arr[ind-1]) { 


                r = ind;
            }
            
            maxDist = Math.max(maxDist, ind - l + 1);
            
            ind++;
        }

        //7 7 7 7 7 8 8 8 8 7 7 7 7 6 6 6 6
        
        System.out.println(maxDist);
    }
}
