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.HashMap;
import java.util.Map;
import java.util.Scanner;

/**
 *
 * @author thachlp
 */
public class Main {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextInt();
        }
        System.out.println(process(arr, n));

    }

    static int process(int[] a, int n) {
        int count = 0, result = 0;
        Map<Integer, Integer> map = new HashMap<>();
        int l = 0;
        for (int i = 0; i < n; i++) {
            Integer value = map.get(a[i]);
            if (value == null) {
                map.put(a[i], 1);
            } else {
                map.put(a[i], ++value);
            }
            count++;
            while (map.size() > 2) {
                Integer tmp = map.get(a[l]);
                if (tmp > 1) {
                    map.put(a[l], --tmp);
                } else {
                    map.remove(a[l]);
                }
                count--;
                l++;
            }
            if (count > result) {
                result = count;
            }
        }
        return result;
    }
}