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.ArrayList;
import java.util.List;
import java.util.Scanner;


public class bai2_them {

    public static void main(String[] args) {

        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(scanner.nextInt());
        }

        List<Integer> counterList = new ArrayList<>();
        for (int i = 0; i < 100001; i++) {
            counterList.add(0);
        }
        int i = 0;
        int j = 0;
        int max = 0;
        int count = 0;


        while (i < n && j < n) {
            if (count <= 2) {
                int value = counterList.get(list.get(i));
                if (value == 0) {
                    count++;
                }
                counterList.set(list.get(i), value + 1);
                if (count <= 2) {
                    max = max < i - j + 1 ? i - j + 1 : max;
                }
                i++;
            } else {
                int value = counterList.get(list.get(j));
                if (value == 1) {
                    count--;
                }
                counterList.set(list.get(j), value - 1);
                j++;
            }

        }

        System.out.print(max);
    }
}