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.PrintWriter;
import java.util.StringTokenizer;

public class B {
    BufferedReader br;
    PrintWriter out;
    StringTokenizer stok;

    String nextToken() throws IOException {
        while (stok == null || !stok.hasMoreTokens()) {
            String s = br.readLine();
            if (s == null) {
                return null;
            }
            stok = new StringTokenizer(s);
        }
        return stok.nextToken();
    }

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

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

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

    char nextChar() throws IOException {
        return (char) (br.read());
    }

    String nextLine() throws IOException {
        return br.readLine();
    }

    boolean check(int[] a, int m, int n) {
        for (int i = 0; i < n - m + 1; i++) {
            if (max.minPos(i, i + m - 1) - min.minPos(i, i + m - 1) <= 1) {
                return true;
            }
        }
        return false;
    }
    RmqSparseTableMin min;
    RmqSparseTableMax max;
    private void solve() throws IOException {
        int n = nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        int l = 1;
        int r = n + 1;
        min = new RmqSparseTableMin(a);
        max = new RmqSparseTableMax(a);
        while (r - l > 1) {
            int m = (l + r) / 2;
            if (check(a, m, n)) {
                l = m;
            } else {
                r = m;
            }
        }
        out.println(l);
    }

    class RmqSparseTableMin {

        int[] logTable;
        int[][] rmq;
        int[] a;

        public RmqSparseTableMin(int[] a) {
            this.a = a;
            int n = a.length;

            logTable = new int[n + 1];
            for (int i = 2; i <= n; i++)
                logTable[i] = logTable[i >> 1] + 1;

            rmq = new int[logTable[n] + 1][n];

            for (int i = 0; i < n; ++i)
                rmq[0][i] = i;

            for (int k = 1; (1 << k) < n; ++k) {
                for (int i = 0; i + (1 << k) <= n; i++) {
                    int x = rmq[k - 1][i];
                    int y = rmq[k - 1][i + (1 << k - 1)];
                    rmq[k][i] = a[x] <= a[y] ? x : y;
                }
            }
        }

        public int minPos(int i, int j) {
            int k = logTable[j - i];
            int x = rmq[k][i];
            int y = rmq[k][j - (1 << k) + 1];
            return a[x] <= a[y] ? a[x] : a[y];
        }
    }

    class RmqSparseTableMax {

        int[] logTable;
        int[][] rmq;
        int[] a;

        public RmqSparseTableMax(int[] a) {
            this.a = a;
            int n = a.length;

            logTable = new int[n + 1];
            for (int i = 2; i <= n; i++)
                logTable[i] = logTable[i >> 1] + 1;

            rmq = new int[logTable[n] + 1][n];

            for (int i = 0; i < n; ++i)
                rmq[0][i] = i;

            for (int k = 1; (1 << k) < n; ++k) {
                for (int i = 0; i + (1 << k) <= n; i++) {
                    int x = rmq[k - 1][i];
                    int y = rmq[k - 1][i + (1 << k - 1)];
                    rmq[k][i] = a[x] >= a[y] ? x : y;
                }
            }
        }

        public int minPos(int i, int j) {
            int k = logTable[j - i];
            int x = rmq[k][i];
            int y = rmq[k][j - (1 << k) + 1];
            return a[x] >= a[y] ? a[x] : a[y];
        }
    }

    void run() throws IOException {
        //br = new BufferedReader(new FileReader("input.txt"));
        //out = new PrintWriter("output.txt");
         br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        br.close();
        out.close();
    }

    public static void main(String[] args) throws IOException {
        // Locale.setDefault(Locale.US);
        new B().run();
    }
}