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

public class B333 {
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

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

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

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

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

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	public static void main(String[] args) {
		FastReader fr = new FastReader();
		int n = fr.nextInt();
		int[] arr = new int[n];
		for (int i = 0; i < n; i++) {
			arr[i] = fr.nextInt();
		}
		int maxidx = 0;
		int minidx = 0;
		int maxval = Integer.MIN_VALUE;
		int val = 0;
		int temp = 0;
		for (int i = 0; i < arr.length - 1;) {
			while (i < arr.length - 1 && arr[i] == arr[i + 1]) {
				i++;
				val++;
				maxidx++;
				minidx++;
			}
			if (i == arr.length - 1) {
				if (val > maxval) {
					maxval = val;
					temp = Math.max(maxidx, minidx) - 1;
				}
				break;
			}
			if (arr[i + 1] > arr[i]) {
				maxidx = i + 1;
				minidx = i;
			} else {
				maxidx = i;
				minidx = i + 1;
			}
			while ((arr[i] == arr[minidx] || arr[i] == arr[maxidx]) && i < arr.length - 1) {
				if (arr[i] == arr[minidx]) {
					minidx = i;
				} else {
					maxidx = i;
				}
				val++;
				i++;
			}
			if (val >= maxval) {
				maxval = val;
				temp = Math.max(maxidx, minidx);
			}
			val = 0;
			if (i == arr.length - 1) {
				break;
			}
			if (arr[i] > arr[maxidx]) {
				i = minidx + 1;
			} else if (arr[i] < arr[minidx]) {
				i = maxidx + 1;
			}
		}
		if ((arr[n - 1] == arr[minidx] || arr[n - 1] == arr[maxidx]) && (temp == n - 2 || temp == n - 1)) {
			maxval++;
		} 

		System.out.println(maxval);

	}
}