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 sc = new Scanner(System.in);

        int n = sc.nextInt();
        int[] massive = new int[n];

        for (int i = 0; i<n; i++) {
            massive[i] = sc.nextInt();
        }

        int max_length = 1;
        int length = 1;
        int l = 0;
        int r = 0;
        int max = massive[0];
        int min = massive[0];
        int balance = 0;

        for (int i = 1; i<n; i++) {
            max = Math.max(max, massive[i]);
            min = Math.min(min, massive[i]);
            balance=max - min;
            r = i;
            if (balance <=1 ) {
                length++;
            } else {
                max_length = Math.max(max_length, length);
                length = 1;
                l = i;
                max = massive[i];
                min = massive[i];
                while (Math.abs(massive[i]-massive[l-1])<=1) {
                    l--;
                    max = Math.max(max, massive[l]);
                    min = Math.min(min, massive[l]);
                    length++;
                }
                balance = massive[r]-massive[l];
            }
        }
        max_length = Math.max(max_length, length);


        System.out.println(max_length);
    }
}