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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.StringTokenizer;

import static java.lang.Math.abs;

/**
 * 602B
 *
 * @author artyom
 */
public class ApproximatingAConstantRange implements Runnable {
    private BufferedReader in;
    private PrintStream out;
    private StringTokenizer tok;

    private void solve() throws IOException {
        int n = nextInt(), maxLen = 0, si = 0;
        int[] a = readIntArray(n);
        for (int i = 1, sj = -1, mm = a[0], mx = a[0]; i < n; i++) {
            if (abs(a[i] - a[i - 1]) == 1) {
                if (mm != mx && (a[i] < mm || a[i] > mx)) {
                    if (i - si > maxLen) {
                        maxLen = i - si;
                    }
                    si = sj;
                }
                sj = i;
                if (a[i] < mm) {
                    mx = mm;
                    mm = a[i];
                } else if (a[i] > mx) {
                    mm = mx;
                    mx = a[i];
                }
            }
        }
        if (n - si > maxLen) {
            maxLen = n - si;
        }
        out.print(maxLen);
    }

    //--------------------------------------------------------------
    public static void main(String[] args) {
        new ApproximatingAConstantRange().run();
    }

    @Override
    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = System.out;
            tok = null;
            solve();
            in.close();
        } catch (IOException e) {
            System.exit(0);
        }
    }

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

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

    private int[] readIntArray(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = nextInt();
        }
        return arr;
    }
}