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.util.*;
public class Main {
static int N = 100000;
    static int n, q, v[] = new int[N], p[] = new int[N], d[] = new int[N],dp2[][] = new int[N][17], dp[][] = new int[N][17];
    static int getMin(int a, int b){
        int log = (int)(Math.log(b-a+1)/Math.log(2));
        return Math.min(dp[a][log], dp[b - (1 << log) + 1][log]);
    }

    static int RMQ(int a, int b)
    {
        int k = (int)(Math.log(b-a+1)/Math.log(2));
        return Math.max(dp2[a][k], dp2[b+1-(int)Math.pow(2, k)][k]);
    }

    public static void main (String []args)  {
    Scanner s =new Scanner (System.in);
    int M=Integer.MIN_VALUE,m=Integer.MAX_VALUE;
    
    int n=s.nextInt();
    for (int i = 0; i<n; ++i)
        v[i]=s.nextInt();
        for (int i = 0; i<n; ++i)
            dp[i][0] = v[i];
        for (int k = 1; (1 << k) <= n; ++k)
            for (int i = 0; i + (1 << k) <= n; ++i)
                dp[i][k] = Math.min(dp[i][k - 1], dp[i + (1 << (k - 1))][k - 1]);
        for (int i = 0; i<n; ++i)
            dp2[i][0] = v[i];
        for (int k = 1; (1 << k) <= n; ++k)
            for (int i = 0; i + (1 << k) <= n; ++i)
                dp2[i][k] = Math.max(dp2[i][k - 1], dp2[i + (1 << (k - 1))][k - 1]);
        
        int r=0,l=0;
        int ans=1;
        while (r<n && l<n)
        {
            int mx=RMQ(l, r),mn=getMin(l, r);
            if ( mx - mn <=1)
            {
                ans=Math.max(ans, r-l+1);
                r++;
            } else {
                ans=Math.max(ans, r-l);
                l++;
            }
        }
        System.out.println(ans);

    s.close();
    }}
