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 B {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int[] ar = new int[n];
		for (int i = 0; i < ar.length; i++) {
			ar[i] = in.nextInt();
		}
		int min = ar[0];
		int max = ar[0];
		int len = 1;
		int clen = 1;
		int eq=0;
		for (int i = 1; i < ar.length; i++) {
			int b = ar[i];
			if (Math.abs(b - min) <= 1 && Math.abs(b - max) <= 1) {
				clen++;
				if (b <= min)
					min = b;
				if (b >= max)
					max = b;
				if (clen > len)
					len = clen;
				// System.out.println(clen);
			} else {
				if (Math.abs(b - ar[i - 1]) <= 1 && Math.abs(b - ar[i - 1]) <= 1) {
					clen =1+eq;
					min = Math.min(b, ar[i - 1]);
					max = Math.max(b, ar[i - 1]);
				} else {
					clen = 1;
					min = b;
					max = b;
				}

			}
			if(b==ar[i-1])
				eq++;
			else
				eq=1;
			// System.out.println(clen+" **");
		}
		if (clen > len)
			len = clen;
		System.out.println(len);
	}
}
