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.*;

/* Mighty Cohadar */
public class Bravo {

	final int n;
	final int[] A;
	int l;
	int r;
	int min;
	int max;

	public Bravo(int n, int[] A) {
		this.n = n;
		this.A = A;
		this.min = A[0];
		this.max = A[0];
	}

	public void goback() {
		while (l > 0) {
			if (A[l - 1] == min || A[l - 1] == max) {
				l--;
			} else if (min == max) {
				if (Math.abs(A[l - 1] - min) <= 1) {
					min = Math.min(min, A[l - 1]);
					max = Math.max(max, A[l - 1]);
				} else {
					break;
				}
			} else {
				break;
			}
		}
	}

	public int solve() {
		int res = 1;
		while (r < n - 1) {
			r++;
			if (A[r] != min && A[r] != max) {
				l = r;
				min = A[r];
				max = A[r];
				goback();
			}
			res = Math.max(res, r - l + 1);
		}
		return res;
	}
	

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		assert 2 <= n && n <= 1e5 : "out of range, n: " + n;
		int[] A = scanArray(scanner, n);
		Bravo o = new Bravo(n, A);
		System.out.println(o.solve());
	}

	static int[] scanArray(Scanner scanner, int n) {
		int[] A = new int[n];
		for (int i = 0; i < n; i++) {
			A[i] = scanner.nextInt();
			assert 1 <= A[i] && A[i] <= 1e5 : "out of range, A[i]: " + A[i];
		}
		return A;
	}

}

