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 Approximating {

	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		int[] nums = new int[n];
		for(int i = 0; i < n; i++)
			nums[i] = s.nextInt();
		int j = 1, i = 0;
		int max = 0;
		int mx = 0, mn = 100000;
		int mnIdx = 0, mxIdx = 0;
		for(i = 0; i < n; i++) {
			if(j >= n)
				break;
			mx = Math.max(nums[i], nums[j]);
			mn = Math.min(nums[i], nums[j]);
			if(mnIdx >= i && mnIdx <= j)
				mn = nums[mnIdx];
			if(mxIdx >= i && mxIdx <= j)
				mx = nums[mxIdx];

			while(mx - mn <= 1) {
				j++;
				if(j >= n)
					break;
				if(nums[j] >= mx) {
					mx = nums[j];
					mxIdx = j;
				}
				if(nums[j] <= mn) {
					mn = nums[j];
					mnIdx = j;
				}
			}
			max = Math.max(max, j - i);
		}
		System.out.print(max);
	}

}
