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.Scanner;
public class Main {
public static void main(String args[]) {
   Scanner java=new Scanner(System.in);
int i,n,c,m,e,d;
n=java.nextInt();
int a[]=new int[n];
for(i=0;i<n;i++)
{
a[i]=java.nextInt();
}
i=1;
c=1;
m=1;
e=1;
d=0;
while(i<n&&a[i]==a[i-1]) i++;
if(i<n){c=c+i; d=a[i]-a[i-1];}
else c=n;
if(c>m) m=c;
i++;
while(i<n)
{
if(a[i]-a[i-1]!=d)
{
c++;
if(a[i]-a[i-1]!=0) {d=a[i]-a[i-1]; e=1;}
else e++;
}
else
{
    c=e+1;
    e=1;
    
}
if(c>m) m=c;
i++;
}
System.out.println(m);
}
}