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.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;
import java.util.concurrent.CountDownLatch;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.TimeUnit;

public class Codeforces implements Runnable {

    BufferedReader in;
    PrintWriter out;
    StringTokenizer tok = new StringTokenizer("");

    void init() throws FileNotFoundException {
        //in = new BufferedReader(new FileReader("dfsongrid.in"));
        //out = new PrintWriter("dfsongrid.out");
        try {
            in = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter("output.txt");
        } catch (Throwable e) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        }
    }

    String readString() throws IOException {
        while (!tok.hasMoreTokens()) {
            try {
                tok = new StringTokenizer(in.readLine());
            } catch (Exception e) {
                return null;
            }
        }
        return tok.nextToken();
    }

    int readInt() throws IOException {
        return Integer.parseInt(readString());
    }

    int[] readIntArray(int size) throws IOException {
        int[] res = new int[size];
        for (int i = 0; i < size; i++) {
            res[i] = readInt();
        }
        return res;
    }

    long readLong() throws IOException {
        return Long.parseLong(readString());
    }

    double readDouble() throws IOException {
        return Double.parseDouble(readString());
    }

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

    long timeBegin, timeEnd;

    void time() {
        timeEnd = System.currentTimeMillis();
        System.err.println("Time = " + (timeEnd - timeBegin));
    }

    long memoryTotal, memoryFree;

    void memory() {
        memoryFree = Runtime.getRuntime().freeMemory();
        System.err.println("Memory = " + ((memoryTotal - memoryFree) >> 10)
                + " KB");
    }

    public void run() {
        try {
            timeBegin = System.currentTimeMillis();
            memoryTotal = Runtime.getRuntime().freeMemory();
            init();
            solve();
            out.close();
            if (System.getProperty("ONLINE_JUDGE") == null) {
                time();
                memory();
            }
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(-1);
        }
    }

    boolean check(int[] a, int k) {
        int n = a.length;
        TreeMap<Integer, Integer> map = new TreeMap<>();
        for (int i = 0; i < k; i++) {
            Integer inm = map.get(a[i]);
            map.put(a[i], inm == null ? 1 : inm + 1);
        }
        if (map.lastKey() - map.firstKey() <= 1) return true;

        for (int i=k;i<n;i++) {
            Integer inm = map.get(a[i - k]);
            if (inm == 1) {
                map.remove(a[i - k]);
            } else {
                map.put(a[i - k], inm - 1);
            }

            inm = map.get(a[i]);
            map.put(a[i], inm == null ? 1 : inm + 1);
            if (map.lastKey() - map.firstKey() <= 1) return true;
        }
        return false;
    }

    void solve() throws IOException {
        int n = readInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = readInt();
        }


        int l = 1;
        int r = n;
        int ans = -1;

        while (l <= r) {
            int mid = (l  +r) / 2;
            if (check(a, mid)) {
                ans = mid;
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        out.println(ans);
    }
}