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.util.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.Stack;
import java.io.InputStream;
import java.util.Vector;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author Himalay(himalayjoriwal@gmail.com)
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        MaximumXorSecondary solver = new MaximumXorSecondary();
        solver.solve(1, in, out);
        out.close();
    }
}

class MaximumXorSecondary {
    

    int arr[];
    int n;

    public void solve(int testNumber, Scanner in, PrintWriter out) {
        n = in.nextInt();
        arr = new int[n];

        for (int i = 0; i < n; i++)
            arr[i] = in.nextInt();
        out.println(Math.max(func(arr), func(reverse(arr))));
    }

    private int func(int[] arr) {
        
        int ans = 0;
        Stack<Integer> st = new Stack<Integer>();
        for (int i = 0; i < n; i++) {
            while (!st.isEmpty() && st.peek() < arr[i])
                st.pop();
            if (!st.isEmpty())
                ans = Math.max(ans, st.peek() ^ arr[i]);
            st.push(arr[i]);
        }
        return ans;
    }

    private int[] reverse(int[] arr) {
        int n = arr.length;
        int ret[] = new int[n];
        for (int i = 0; i < n; i++)
            ret[i] = arr[n-1-i];
        return ret;
    }

}
