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.*;
public class Abc {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int ar[] = new int[n];
		int f [] = new int[100001];
		for(int i = 0; i < n; i++) {
			ar[i] = in.nextInt();
		}
		int start = 0, end = 0;
		int max = 0, min = 1000000;
		int maxLen = 0;
		while(end < n) {
			f[ar[end]]++;
			//System.out.println("Found : "+ ar[end]);
			if(end == 0) {
				max = min =ar[0];
			} else if(ar[end] > max) {
				//System.out.println(max)
				//System.out.println(ar[end] + " > (Max) " + max);
				max = ar[end];
				if(ar[end] - min > 1) {
					while(true) {
						f[ar[start]] --;
						//System.out.println("start f remaining " + f[ar[start]]);
						if(ar[start] == min && f[min] == 0) {
							//System.out.println("min : " + min + " f[min] : "+ f[min]);
							min ++;
							start++;
							maxLen = Math.max(maxLen, end - start + 1);
							break;
						}
						start++;
					}
				}
			} else if(ar[end] < min) {
				//System.out.println(ar[end] + " < (Min) " + max);
				min = ar[end];
				if(max - ar[end] > 1) {
					while(true) {
						f[ar[start]] --;
						if(ar[start] == max && f[max] == 0) {
							max --;
							start++;
							maxLen = Math.max(maxLen, end - start + 1);
							break;	
						}
						start++; 
					}
				}
			}
			if(max - min <= 1) {
				maxLen = Math.max(maxLen, end - start + 1);
			}
			end ++;
		}
		System.out.println(maxLen);
	}
}