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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.List;
import java.util.Scanner;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Oleksii Sosevych (alexey_sosevich@ukr.net)
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskB {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int min = in.nextInt();
            int max = min;
            int minIndex = 0;
            int maxIndex = 0;
            int maxLen = 0;
            List<Integer> list = new ArrayList<>();
            list.add(min);
            for (int i = 1; i < n; i++) {
                int currentVal = in.nextInt();
                list.add(currentVal);
                max = Math.max(currentVal, max);
                min = Math.min(currentVal, min);
                if (Math.abs(max - min) <= 1) {
                    maxLen = Math.max(maxLen, i - minIndex + 1);
                } else {
                    max = currentVal;
                    min = currentVal;
                    minIndex = i;
                    while (Math.abs(list.get(minIndex - 1) - min) <= 1 && Math.abs(list.get(minIndex - 1) - max) <= 1) {
                        min = Math.min(list.get(minIndex - 1), min);
                        max = Math.max(list.get(minIndex - 1), max);
                        minIndex--;
                    }
                }
            }
            out.println(maxLen);
        }

    }
}
