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.Arrays;
import java.util.Scanner;
import java.io.BufferedReader ; 
import java.io.InputStreamReader ; 
import java.io.IOException  ;
public class B602_ApproxmatingaConstantRange {
	 
	public static void main(String[] args)throws IOException  {
		Scanner ss = new Scanner (new BufferedReader(new InputStreamReader(System.in))) ; 
		int n  = ss.nextInt() ; 
		int [] a = new int [n] ; 
		int c = 1 ; a[0] = ss.nextInt() ; 
		int max = a[0] ; 
		int min = a[0] ; 
		int lon = 0 ; 
	//	System.out.println("max 	min 	i 	c ");
		for (int i = 1 ; i <a.length ; i ++  ) {
		//	System.out.println(max + "	" + min + "	" + i + "	" + c );
			a[i] = ss.nextInt() ; 
			if (max < a[i])
				max = a[i] ; 
			else if (min > a[i] ) 
				min = a[i] ; 
			if ((int ) (Math.abs(max-min)) <=  1 ) { 
				c++ ; 
			}
			else { 
				if (c>lon)
					lon = c ; 
				if ((int ) Math.abs(a[i] -a[i-1]) <=1){
					c= 2 ; 
					if (a[i-1] < a[i]) {
						max = a[i] ;  
						min = a[i-1] ; } 
					
					else {
						max = a[i-1] ; 
						min = a[i] ; 
					}
					for (int j = i-2 ; j >= 0 ; j -- ) { 
						if (a[j] == a[j+1])
							 c++ ; 
						else 
							break ; 
					}
				}
				else {
					c = 1 ; 
					min = a[i] ; 
					max = a[i] ; 
				}

					
			}
			
		}
		if (c>lon)
			lon = c ; 
		System.out.println(lon);
	}

}
