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].
// package CF;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class B602 {

	public static void main(String[] args) throws IOException {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int arr[] = new int[n];
		for (int i = 0; i < n; i++)
			arr[i] = in.nextInt();
		int cnt[] = new int[(int) 1e6 + 5];
		TreeSet<Integer> ts = new TreeSet<>();
		cnt[arr[0]]++;
		ts.add(arr[0]);
		int res = 0;
		for (int l = 0, r = 1; l < n; l++) {
			while (r < n) {
				int max = Math.max(arr[r], ts.last());
				int min = Math.min(arr[r], ts.first());
				if(max - min > 1)
					break;
				ts.add(arr[r]);
				cnt[arr[r]]++;
				r++;
			}
//			System.err.println(l + " " + r);
//			System.err.println(ts.toString());
			res = Math.max(res, r - l);
			cnt[arr[l]]--;
			if(cnt[arr[l]] == 0)
				ts.remove(arr[l]);
		}
		System.out.println(res);

	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}
	}
}
// 