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.io.*;
import java.util.*;

public class P2 {
	public static void main (String[] args) {
		Scanner sc = new Scanner (System.in);
		PrintStream op = System.out;
		int n = sc.nextInt();
		int[] a = new int [n];
		for (int i = 0; i < n; i++)
			a[i] = sc.nextInt();
		int last = 0, max = 0, lastChange = 0;
		boolean isUp = false, started = false;
		for (int i = 1; i < n; i++) {
			if (a[i] > a[i - 1]) {
				if (started && isUp) {
					max = Math.max(max, i - last);
					last = lastChange;
				}
				lastChange = i;
				isUp = true;
				started = true;
			}
			else if (a[i] < a[i - 1]) {
				if (started && !isUp) {
					max = Math.max(max, i - last);
					last = lastChange;
				}
				lastChange = i;
				isUp = false;
				started = true;
			}
		}
		max = Math.max(max, n - last);
		op.println(max);
		sc.close();
		op.close();
	}
}
