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.awt.List;
import java.awt.Point;
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.NavigableMap;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeMap;

public class graphs {
	static Scanner in = new Scanner();
	static PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) throws NumberFormatException, IOException {
		int n = in.nextInt();
		int arr[] = new int[n];
		for (int i = 0; i < n; arr[i++] = in.nextInt())
			;
		Point[] min_max = new Point[n];
		for (int i = 0; i < n; i++) {
			min_max[i] = new Point();
		}
		int dp[] = new int[n];
		dp[0] = 1;
		min_max[0] = new Point(0, 0);
		for (int i = 1; i < n; i++) {
			if (Math.abs(arr[min_max[i - 1].x] - arr[i]) <= 1 && Math.abs(arr[min_max[i - 1].y] - arr[i]) <= 1) {
				dp[i] = dp[i - 1] + 1;
				int min_i = arr[min_max[i - 1].x] > arr[i] ? i : min_max[i - 1].x;
				int max_i = arr[min_max[i - 1].y] > arr[i] ? min_max[i - 1].y : i;
				min_max[i] = new Point(min_i, max_i);
			} else {
				int now_i = i;
				min_max[i] = new Point(i, i);
				while (now_i >= 0 && Math.abs(arr[i] - arr[now_i]) <= 1) {
					int min_i = arr[min_max[i].x] > arr[now_i] ? now_i : min_max[i].x;
					int max_i = arr[min_max[i].y] > arr[now_i] ? min_max[i].y : now_i;
					min_max[i].x = min_i;
					min_max[i].y = max_i;
					now_i--;
				}
				dp[i] = i - now_i;
			}
		}
		int max = Integer.MIN_VALUE;
		for (Integer i : dp) {
			// out.print(i+" ");
			max = Math.max(max, i);
		}
		out.println(max);
		out.flush();
		out.close();

	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		Scanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		Scanner(String file) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(file));
		}

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

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

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

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

	}

}
