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

public class B602 {
	public static void main(String[] args) throws Exception {
		int n = i(), a[] = new int[n];

		for (int i = 0; i < n; i++) {
			a[i] = i();
		}

		int ans = 0;
		TreeMap<Integer, Integer> c = new TreeMap<>();

		int l = 0, r = 0;
		while (l <= r && r < n) {
			c.compute(a[r], (k, v) -> v == null ? 1 : v + 1);
			r++;

			while (l <= r && c.lastKey() - c.firstKey() > 1) {
				c.compute(a[l], (k, v) -> v == 1 ? null : v - 1);
				l++;
			}

			ans = Math.max(ans, r - l);
		}

		out.println(ans);
		
		out.close();
	}

	static BufferedReader in;
	static StringTokenizer st = new StringTokenizer("");
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
	static {
		try {
			in = new BufferedReader(new FileReader("cf.in"));
		} catch (Exception e) {
			in = new BufferedReader(new InputStreamReader(System.in));
		}
	}
	static void check() throws Exception {
		while (!st.hasMoreTokens()) st = new StringTokenizer(in.readLine());
	}
	static int i() throws Exception {check(); return Integer.parseInt(st.nextToken());}
	static String s() throws Exception {check(); return st.nextToken();}
	static double d() throws Exception {check(); return Double.parseDouble(st.nextToken());}
	static long l() throws Exception {check(); return Long.parseLong(st.nextToken());}
}
