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 Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
 
        int n = scanner.nextInt();
        int[] a = new int[n];
 
        for (int i = 0; i < n; i++) {
            a[i] = scanner.nextInt();
        }
 
        int maxLength = 0;
 
        // Solution
        int beginIndex = 0;
        while (beginIndex < n - 1) {
            int begin = beginIndex;
            int end = beginIndex + 1;
            int max = a[beginIndex];
            int maxIndex = beginIndex;
            int min = a[beginIndex];
            int minIndex = beginIndex;
            for (int j = beginIndex; j < n; j++) {
                if (a[j] > max) {
                    max = a[j];
                    maxIndex = j;
                }
 
                if (a[j] < min) {
                    min = a[j];
                    minIndex = j;
                }
 
                if (max - min > 1) {
                    end = j - 1;
                    break;
                } else {
                    beginIndex = maxIndex > minIndex ? maxIndex : minIndex;
                }
 
                if (j == n - 1) {
                    end = j;
                }
            }
 
            if (end - begin + 1 > maxLength) {
                maxLength = end - begin + 1;
            }
 
            if (maxLength == n || end == n - 1) {
                break;
            }
 
            if (beginIndex <= begin) {
                beginIndex++;
            }
        }
 
        System.out.println(maxLength);
    }
}