Bike loves looking for the second maximum element in the sequence. The second maximum element in the sequence of distinct numbers x1, x2, ..., xk (k > 1) is such maximum element xj, that the following inequality holds: <image>.

The lucky number of the sequence of distinct positive integers x1, x2, ..., xk (k > 1) is the number that is equal to the bitwise excluding OR of the maximum element of the sequence and the second maximum element of the sequence.

You've got a sequence of distinct positive integers s1, s2, ..., sn (n > 1). Let's denote sequence sl, sl + 1, ..., sr as s[l..r] (1 ≤ l < r ≤ n). Your task is to find the maximum number among all lucky numbers of sequences s[l..r].

Note that as all numbers in sequence s are distinct, all the given definitions make sence.

Input

The first line contains integer n (1 < n ≤ 105). The second line contains n distinct integers s1, s2, ..., sn (1 ≤ si ≤ 109).

Output

Print a single integer — the maximum lucky number among all lucky numbers of sequences s[l..r].

Examples

Input

5
5 2 1 4 3


Output

7


Input

5
9 8 3 5 7


Output

15

Note

For the first sample you can choose s[4..5] = {4, 3} and its lucky number is (4 xor 3) = 7. You can also choose s[1..2].

For the second sample you must choose s[2..5] = {8, 3, 5, 7}.
import java.io.*;
import java.util.*;

public class DSol {
    
    int go (int[] a, int n) {
        int max = Integer.MIN_VALUE;
        Stack<Integer> sta = new Stack<Integer>();
        sta.push(a[0]);
        
        for (int i = 1; i < n; ++i) {
            int cur = a[i];
            while (!sta.empty() && cur > sta.peek()) {
                sta.pop();
            }
            if (!sta.empty()) {
                max = Math.max(max, cur ^ sta.peek());
            }
            sta.push(a[i]);
        }
        
        return max;
    }
    
    public void solve() throws IOException {
        int n = nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; ++i) {
            a[i] = nextInt();
        }
        int[] b = new int[n];
        int j = 0;
        for (int i = n - 1; i >= 0; --i) {
            b[i] = a[j];
            j++;
        }
        out.println(Math.max(go(a, n), go(b, n)));
    }

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

    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader((System.in)));
            out = new PrintWriter(System.out);

            st = null;
            rnd = new Random();

            //double start = System.currentTimeMillis();
            solve();
            //double total = (System.currentTimeMillis() - start) / 1000.0;
            //total *= 10;

            //out.println("Total time: " + total + " sec.");

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

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

        return st.nextToken();
    }

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

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

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

    private BufferedReader in;
    private PrintWriter out;
    private StringTokenizer st;
    private Random rnd;
}