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.lang.*;
import java.io.*;
import java.util.*;


public class Range {
	
	public static void main(String[] args) throws java.lang.Exception {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskA solver = new TaskA();
		solver.solve(in, out);
		out.close();
	}
}

class TaskA {
	public final static int maxn = 100005;
	int[] pos = new int[maxn];
	
	public void solve(InputReader in, PrintWriter out) {
		int n = in.nextInt();
		int[] a = new int[n];
		int i, j, k = 0;
		int mn, mx;
		int ans = 0;
		
		for (i=0; i<n; ++i) a[i] = in.nextInt();
		
		i = 0;
		pos[a[i]] = i;
		while (i<n && a[i]==a[0]) ++i;
		if (i >= n) {
			out.println(n);
			return ;
		}
		mn = mx = a[0];
		while (i < n) {
			// out.println("mn = " + mn + ", mx = " + mx + ", k = " + k + ", i = " + i);
			while (true) {
				// out.println("mn = " + mn + ", mx =  " + mx + ", i = " + i);
				mx = Math.max(mx, a[i]);
				mn = Math.min(mn, a[i]);
				j = i;
				pos[a[i]] = i;
				while (i<n && a[i]==a[j]) ++i;
				if (i >= n) break;
				if (Math.max(mx, a[i]) - Math.min(mn, a[i]) > 1) {
					break;
				}
			}
			ans = Math.max(ans, i-k);
			k = pos[a[i-1]];
			mn = mx = a[i-1];
		}
		
		out.println(ans);
	}
} 

class InputReader {
	public BufferedReader reader;
	public StringTokenizer tokenizer;
	
	public InputReader(InputStream stream) {
		reader = new BufferedReader(new InputStreamReader(stream), 32768);
		tokenizer = null;
	}
	
	public String next() {
		while (tokenizer==null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}
	
	public int nextInt() {
		return Integer.parseInt(next());
	}
	
	public long nextLong() {
		return Long.parseLong(next());
	}
}