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 {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
    Scanner s = new Scanner(System.in) ;
    int n = s.nextInt() ;
    int ar[] = new int[n] ;
    
    int i = 0 , counter = 0 ;
   for(int j = 0 ; j < n ; j++)
       ar[j] = s.nextInt() ;
    int a1 = ar[0] ;
    
    while(a1 == ar[i])
    {
    i++;
    counter ++ ;
    if(i == n)
        break ;
        
    }
       
    int a2 ;
    if(i < n ) 
    { a2 = ar[i] ; 
        
    while(a1 == ar[i] || a2 == ar[i])
    {
        i++ ;
        counter++;
      if(i == n)
        break ;
    }
    
    }
    
    
    
        int max = counter;
    if(max == n)
    {System.out.println(max);
     System.exit(0);
    }
    while(i < n - 1){
    a1 = ar[i - 1] ;
    a2 = ar[i] ;
  
    int j = i - 2 ;
    counter = 2 ;    
    while (a1 == ar[j] || a2 == ar[j]) {
            j--;
            counter++;
            
        }
    if(i < n - 1)
    {i++;
    while(a1 == ar[i] || a2 == ar[i])
    {
    i++;
    counter++;
    if(i == n)
        break ;
    }
    
    
    }
        
    if(max < counter)
        max = counter ;
    }
        System.out.println(max);
    }
    
}
