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.Scanner;

public class BigO1806 {

	public static void main(String[] args) {
		Scanner UserInput = new Scanner(System.in);
		int num = UserInput.nextInt();
		int[] list = new int[num];
		for (int i = 0; i < num; i++)
			list[i] = UserInput.nextInt();
		int start = 0;
		int min = 0;
		int max = 0;
		int minP = 0;
		int maxP = 0;
		int maxL = 0;
		int i = 1;
		for (; i < num; i++) {
			maxL++;
			if (list[i] > list[0]) {
				maxP = i;
				minP = i - 1;
				min = list[0];
				max = list[i++];
				break;
			} else if (list[i] < list[0]) {
				minP = i;
				maxP = i - 1;
				max = list[0];
				min = list[i++];
				break;
			}
		}
		for (; i < num; i++) {
			if (list[i] < min) {
				max = min;
				min = list[i];
				start = maxP + 1;
				maxP = minP;
				minP = i;
				if (i - start > maxL)
					maxL = i - start;
			} else if (list[i] == min) {
				minP = i;
				if (i - start > maxL)
					maxL = i - start;
			}
			else if (list[i] == max) {
				maxP = i;
				if (i - start > maxL)
					maxL = i - start;
			}
			else if (list[i] > max) {
				min = max;
				max = list[i];
				start = minP + 1;
				minP = maxP;
				maxP = i;
				if (i - start > maxL)
					maxL = i - start;
			}
		}
		System.out.println(maxL + 1);
		UserInput.close();
	}

}