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.*;
import java.util.StringTokenizer;

public class B {

    public static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    public static StringTokenizer tok;

    public static int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public static long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    public static double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    public static String nextToken() throws IOException {
        if (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    public static void main(String[] args) throws IOException {
        int n = nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }

        int start = 0;
        int result = 1;
        int minimum = a[0];
        int maximum = a[0];
        int minIndex = 0;
        int maxIndex = 0;
        for (int i = 1; i < n; i++) {
            if (a[i] == minimum) {
                minIndex = i;
            }
            if (a[i] == maximum) {
                maxIndex = i;
            }
            if (a[i] != minimum && a[i] != maximum) {
                if (maximum != minimum) {
                    result = Math.max(result, i - start);
                }
                if (a[i] > maximum) {
                    if (maximum != minimum) {
                        start = minIndex + 1;
                    }
                    minimum = maximum;
                    minIndex = maxIndex;
                    maximum = a[i];
                    maxIndex = i;
                } else {
                    if (maximum != minimum) {
                        start = maxIndex + 1;
                    }
                    maximum = minimum;
                    maxIndex = minIndex;
                    minimum = a[i];
                    minIndex = i;
                }
            }
//            if (a[i] != a[i - 1]) {
//                if (minimum == maximum && a[i] < minimum) {
//                    minimum = a[i];
//                } else if (minimum == maximum && a[i] > maximum) {
//                    maximum = a[i];
//                } else if (minimum != maximum) {
//
//                }
//
//            }
        }
        result = Math.max(result, n - start);
        System.out.println(result);
    }
}