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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;

public class Main{

	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
		
		int n = Integer.parseInt(br.readLine()), left = 1, right = 1, nextLeft = 0, result = 1;
		int[] input_nums = new int[n + 1];
		String[] input_string = br.readLine().split(" ");
		
		for(int i = 1; i <= n; i++)
		{
			input_nums[i] = Integer.parseInt(input_string[i - 1]);
		}
		
		while_break :
		while(right <= n)
		{
			while(input_nums[left] == input_nums[right]) 
			{
				if(right == n)
				{
					if(result < right - left + 1)
					{
						result = right - left + 1;
					}
					break while_break;
				}
				right++;
			}
			
			nextLeft = right;
			
			while(input_nums[left] == input_nums[right] | input_nums[nextLeft] == input_nums[right]) 
			{
				if(right == n)
				{
					if(result < right - left + 1)
					{
						result = right - left + 1;
					}
					break while_break;
				}
				right++;
			}		
			
			if(result < right - left)
			{
				result = right - left;
			}
			
			left = nextLeft;
			right = nextLeft + 1;
			nextLeft = 0;
		}
		
		if(nextLeft != 0)
		{
			if(result < n - left + 1)
			{
				result = n - left + 1;
			}
		}
		
		bw.write(result + "\n");
		bw.flush();
	}

}
