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].
// codeforces 602 b 
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class ApproxPostoian {
	
	int n; // number of men
	int []a;  // men
	int maxPost;
	
	void readData(BufferedReader bin) throws IOException {
		// read men
		n = Integer.parseInt(bin.readLine());
		a = new int[n];
		String sss = bin.readLine();
		String []ss = sss.split(" ");
		for (int i=0; i<n; i++) {
			a[i] = Integer.parseInt(ss[i]);
		}		
		bin.close();
	}
	
	
	// we need to use max flow algorithm
	void calculate() {
		int d1 = a[0]-1;
		int d2 = a[0];
		int l1=1, l2=1;
		for (int i=1; i<n; i++) {
			if (a[i]<a[i-1]) {
				// end of d2
				if (maxPost<l2) {
					maxPost=l2;
				}
				d2=a[i];
				l2=l1+1;
				d1=d2-1;
				l1=1;				
			} else if (a[i]>a[i-1]) {
				// end of d1
				if (maxPost<l1) {
					maxPost=l1;
				}
				d1=a[i]-1;
				l1=l2+1;
				d2=d1+1;
				l2=1;				
			} else {
				l1++;
				l2++;
			}
		}
		if (maxPost<l1) {
			maxPost=l1;
		}
		if (maxPost<l2) {
			maxPost=l2;
		}
	}
	
	void writeRes() throws IOException {
		System.out.println(maxPost);
	}

	/**
	 * @param args
	 */
	public static void main(String[] args) throws IOException {
		BufferedReader bin = new BufferedReader(new InputStreamReader(System.in));
		ApproxPostoian p = new ApproxPostoian();
		p.readData(bin);
		p.calculate();
		p.writeRes();
	}
}
