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 BApproximate {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
//		10
//		1 2 2 2 2 3 2 1 1 3

		
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		
		int[] arr = new int[n];
		for(int i = 0; i < n; i++) {
			arr[i] = scan.nextInt();
		}
		
		int idx_max = 0, idx_min = 0, maxLen = 0, len = 0;
		for(int i = 0; i < n; i++) {

			if(arr[i] >= arr[idx_max])
				idx_max = i;
			if(arr[i] <= arr[idx_min])
				idx_min = i;
			int diff = arr[idx_max] - arr[idx_min];
			if(diff > 1) {
				maxLen = Math.max(len, maxLen);
				int j = i;
				while(Math.abs(arr[i] - arr[j]) <= 1 && j >= 0) {
					j--;
				} 
				len = i - j;
				j++;
				if(arr[i] == arr[idx_max])
					idx_min = j;
				if(arr[i] == arr[idx_min])
					idx_max = j;
			}
			else {
				len++;
			}
		}
		
		if(len > maxLen)
			maxLen = len;
		
		System.out.print(maxLen);
			
		
		

		
	}

}
