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;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author MrX
 */
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);
		TaskD solver = new TaskD();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskD {
    public void solve(int testNumber, Scanner in, PrintWriter out) {
        int n = in.nextInt();

        Stack<Long> a = new Stack<Long>();
        long result = 0;
        while (n > 0) {
            long x = in.nextLong();
            while(!a.empty() && a.peek() < x) {
                result = Math.max(result, a.pop() ^ x);
            }
            if (!a.empty()) {
                result = Math.max(result, a.peek() ^ x);
            }

            a.push(x);
            n--;
        }
        out.print(result);
    }
}

