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.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;


public class Constant {

	/**
	 * @param args
	 * @throws IOException 
	 */
	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(in.readLine());
		StringTokenizer st = new StringTokenizer(in.readLine());
		int[] nums = new int[n];
		for(int i = 0; i < n; i++){
			nums[i] = Integer.parseInt(st.nextToken());
		}
		int curInd = 0;
		int maxInd = 0;
		int minInd = 0;
		int cur = nums[0];
		int min = nums[0];
		int max = nums[0];
		int maxRun = 0;
		for(int i = 0; i < n; i++){
			if(Math.abs(max - nums[i]) <= 1 && Math.abs(min - nums[i]) <= 1){
				if(nums[i] > cur && max == cur){
					max = nums[i];
					maxInd = i;
				}
				else if(nums[i] < cur && min == cur){
					min = nums[i];
					minInd = i;
				}
			}
			else{
				int curRun = i - curInd;
				if(curRun > maxRun) maxRun = curRun;
				if(curInd != maxInd && curInd!=minInd){
					i = maxInd < minInd ? maxInd : minInd;
				}
				else if(curInd != maxInd){
					i = maxInd;
				}
				else{
					i = minInd;
				}
				curInd = i;
				maxInd = i;
				minInd = i;
				cur = nums[i];
				max = nums[i];
				min = nums[i];
			}
		}
		int curRun = nums.length - curInd;
		if(curRun > maxRun) maxRun = curRun;
		System.out.println(maxRun);
	}

}
