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.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();

        ArrayList<Integer> arr = new ArrayList<>();
        int value;
        for (int i = 0; i < n; i++) {
            value = sc.nextInt();
            arr.add(value);
        }

        findOut(n, arr);
    }

    private static void findOut(int n, ArrayList<Integer> arr) {
        int count = 0;
        int max = 0;
        int m = arr.get(0);
        int M = arr.get(0);
        int l = 0;
        int r = 0;
        for (; r < n; r++) {
            int value = arr.get(r);

            // recalculate m, M
            if (value > M) M = value;
            if (value < m) m = value;

            if (M - m <= 1) { // thoa dk
                count++;
                max = Math.max(count, max);
            } else {
                M = value;
                m = value;
                for (int i = r - 1; i >= 0; i--) {
                    if (Math.abs(arr.get(i) - value) > 1) {
                        count = r - i;
                        max = Math.max(count, max);

                        // recalculate M, m
                        if (arr.get(i + 1) > M) M = arr.get(i + 1);
                        if (arr.get(i + 1) < m) m = arr.get(i + 1);
                        break;
                    }
                }
            }
        }

        System.out.println(max);
    }
}