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 Main {

	public static void main(String[] args) throws IOException {
		Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		int n = in.nextInt();
		int[] arr = new int[n];
		for(int x = 0; x < n; x++)
			arr[x] = in.nextInt();
		int i = 0, j = 0, s= 0;
		int max = 1;
		for(int x = 1; x < n; x++) {
			if(arr[i] == arr[j] && arr[i]== arr[x]) {
				i++;
				j++;
			}
			else if(arr[x] <= arr[i]) {
				if(arr[j] - arr[x] > 1) {
					max = Math.max(max, x - s);
					s = j + 1;
					j = i;
				}
				i = x;
			}
			else if(arr[x] >= arr[j]) {
				if(arr[x] - arr[i] > 1) {
					max = Math.max(max, x - s);
					s = i + 1;
					i = j;
				}
				j = x;
			}
		}
		max = Math.max(max, n - s);
		System.out.println(max);
	}

}