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 ApproximatingAConstantRange_602B {

	static int[] NN;
	static int[] m;
	static int n;
	static int k;
	
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		n = scan.nextInt();
		NN = new int[n+1];
		m = new int[n+1];
		for(k = 1;k<=n;k++) {
			NN[k] = scan.nextInt();
		}
		System.out.println( solve());
		scan.close();
	}

	public static int solve() {
		int a, ac, b, bc, i, j; // a smaller, b larger, b == a+1
		if (n == 1) return 1;
		for (k = 2; k <= n; k++)
			if (NN[k] != NN[k-1]) break;
		if (k == n+1) return n;
		i = 1; j = k + 1; a = Math.min(NN[k-1], NN[k]); b = Math.max(NN[k-1], NN[k]); m[k] = k;	if (b == NN[k]) {bc = 1; ac = k-1;} else {ac = 1; bc = k - 1;}
	 
		while (true) {
	 
		while (j <= n) {
			if (NN[j] == a) {ac++; m[j] = j-i+1; j++;}
			else if (NN[j] == b) {bc++; m[j] = j-i+1; j++;}
			else {break;}
		}
		if (j == n + 1) break;
		if (NN[j] == a-1) {
			while (bc > 0) {
				if (NN[i] == a) {ac--; i++;}
				else {bc--; i++;}
			}
			b = a; bc = ac; a = a - 1; ac = 0;
		}
		else { // NN[j] == b+1
			while (ac > 0) {
				if (NN[i] == a) {ac--; i++;}
				else {bc--; i++;}
			}
			a = b; ac = bc; b = b + 1; bc = 0;
		}
	 
		}
		int ans = m[1];                             // if (n>5) printf("%d ", m[1]);
		for (k = 2; k <= n; k++) {           // if (n>5) printf("%d ", m[k]);
			ans = Math.max(ans, m[k]); }
		return ans;
	}
}