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.util.*;
import java.io.*;

public class Main {
    public static void main(String[] args) throws IOException {
        InputReader in = new InputReader();
        PrintWriter out = new PrintWriter(System.out);
        Solver s = new Solver();
        int test = 1;
        for (int i = 1; i <= test; i++) {
            s.solve(i, in, out);
        }
        out.close();
    }
}

class Solver {

    int[] tMax, tMin, a;

    void solve(int test_case, InputReader in, PrintWriter out) throws IOException {
        int n = in.nextInt();
        a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
        }
        tMax = new int[4 * n];
        tMin = new int[4 * n];
        build(0, 0, n - 1);
        int low = 1, high = n, ans = 1;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (can(mid)) {
                ans = mid;
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        out.println(ans);
    }

    private boolean can(int k) {
        k--;
        for (int i = 0; i + k < a.length; i++) {
            int max = queryMax(0, 0, a.length - 1, i, i + k);
            int min = queryMin(0, 0, a.length - 1, i, i + k);
//            System.err.println(k + " " + max + " " + min + " " + (i));
            if (max - min <= 1) return true;
        }
        return false;
    }

    private int queryMin(int node, int start, int end, int l, int r) {
        if (start > r || end < l) return Integer.MAX_VALUE;
        if (l <= start && end <= r) return tMin[node];
        int mid = (start + end) / 2;
        int x = queryMin(2 * node + 1, start, mid, l, r);
        int y = queryMin(2 * node + 2, mid + 1, end, l, r);
        return Math.min(x, y);
    }

    private int queryMax(int node, int start, int end, int l, int r) {
        if (start > r || end < l) return 0;
        if (l <= start && end <= r) return tMax[node];
        int mid = (start + end) / 2;
        int x = queryMax(2 * node + 1, start, mid, l, r);
        int y = queryMax(2 * node + 2, mid + 1, end, l, r);
        return Math.max(x, y);
    }

    private void build(int node, int start, int end) {
        if (start == end) {
            tMin[node] = tMax[node] = a[start];
            return;
        }
        int mid = (start + end) / 2;
        build(2 * node + 1, start, mid);
        build(2 * node + 2, mid + 1, end);
        tMax[node] = Math.max(tMax[2 * node + 1], tMax[2 * node + 2]);
        tMin[node] = Math.min(tMin[2 * node + 1], tMin[2 * node + 2]);
    }
}

class InputReader {
    final private int BUFFER_SIZE = 1 << 16;
    private DataInputStream din;
    private byte[] buffer;
    private int bufferPointer, bytesRead;

    public InputReader() {
        din = new DataInputStream(System.in);
        buffer = new byte[BUFFER_SIZE];
        bufferPointer = bytesRead = 0;
    }

    public InputReader(String file_name) throws IOException {
        din = new DataInputStream(new FileInputStream(file_name));
        buffer = new byte[BUFFER_SIZE];
        bufferPointer = bytesRead = 0;
    }

    public String nextLine() throws IOException {
        int SIZE = 1 << 23;
        byte[] buf = new byte[SIZE]; // line length
        int cnt = 0, c;
        while ((c = read()) != -1) {
            if (c == '\n')
                break;
            buf[cnt++] = (byte) c;
        }
        return new String(buf, 0, cnt);
    }

    public int nextInt() throws IOException {
        int ret = 0;
        byte c = read();
        while (c <= ' ')
            c = read();
        boolean neg = (c == '-');
        if (neg)
            c = read();
        do {
            ret = ret * 10 + c - '0';
        } while ((c = read()) >= '0' && c <= '9');

        if (neg)
            return -ret;
        return ret;
    }

    public long nextLong() throws IOException {
        long ret = 0;
        byte c = read();
        while (c <= ' ')
            c = read();
        boolean neg = (c == '-');
        if (neg)
            c = read();
        do {
            ret = ret * 10 + c - '0';
        }
        while ((c = read()) >= '0' && c <= '9');
        if (neg)
            return -ret;
        return ret;
    }

    public double nextDouble() throws IOException {
        double ret = 0, div = 1;
        byte c = read();
        while (c <= ' ')
            c = read();
        boolean neg = (c == '-');
        if (neg)
            c = read();

        do {
            ret = ret * 10 + c - '0';
        }
        while ((c = read()) >= '0' && c <= '9');

        if (c == '.') {
            while ((c = read()) >= '0' && c <= '9') {
                ret += (c - '0') / (div *= 10);
            }
        }

        if (neg)
            return -ret;
        return ret;
    }

    private void fillBuffer() throws IOException {
        bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
        if (bytesRead == -1)
            buffer[0] = -1;
    }

    private byte read() throws IOException {
        if (bufferPointer == bytesRead)
            fillBuffer();
        return buffer[bufferPointer++];
    }

    public void close() throws IOException {
        if (din == null)
            return;
        din.close();
    }

    public String next() throws IOException {
        byte[] buf = new byte[1 << 23]; // line length
        int cnt = 0, c;
        while ((c = read()) != -1) {
            if (c == '\n' || c == ' ')
                break;
            buf[cnt++] = (byte) c;
        }
        return new String(buf, 0, cnt);
    }
}