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.*;
import java.util.*;

public class R333B {
	public static void main(String[] args) throws Exception {
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		Scanner s = new Scanner(System.in);
		
		int n = Integer.parseInt(f.readLine());
		long[] a = new long[n];
		String[] split = f.readLine().split("\\s+");
		for(int i = 0; i < n; i++) a[i] = Long.parseLong(split[i]);
		TreeSet<Long> set = new TreeSet<>();
		long[] freq = new long[100100];
		int i = 0, j = 0;
		int longest = -1;
		while(i < n && j < n) {
			set.add(a[j]);
			freq[(int) a[j]]++;
			long large = set.last(), small = set.first();
//			System.out.printf("%d %d\n", large, small);
			long diff = large-small;
			if(diff <= 1) {
				longest = Math.max(longest, j-i+1);
			} else {
				while(i < n && diff > 1) {
					//remove a[i]
					freq[(int) a[i]]--;
					if(freq[(int) a[i]] == 0) {
						set.remove(a[i]);
					}
					i++;
					diff = set.last() - set.first();
				}
				if(diff <= 1) longest = Math.max(longest, j-i+1);
			}
			j++;
		}
		System.out.println(longest);
	}
}