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.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;
import java.util.Stack;

public class ProblemB {

	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int n = in.nextInt();
		long[] a = new long[n];
		for (int i = 0 ; i < n ; i++) {
			a[i] = in.nextLong();
		}
		
		long[] leftBigger = new long[n];
		long[] rightBigger = new long[n];
		
		// right
		{
			Stack<Integer> si = new Stack<Integer>();
			for (int i = 0 ; i < n ; i++) {
				while (si.size() >= 1 && a[si.peek()] < a[i]) {
					rightBigger[si.pop()] = a[i];
				}
				si.add(i);
			}
			while (si.size() >= 1) {
				rightBigger[si.pop()] = -1;
			}
		}
		
		// left
		{
			Stack<Integer> si = new Stack<Integer>();
			for (int i = n - 1 ; i >= 0 ; i--) {
				while (si.size() >= 1 && a[si.peek()] < a[i]) {
					leftBigger[si.pop()] = a[i];
				}
				si.add(i);
			}
			while (si.size() >= 1) {
				leftBigger[si.pop()] = -1;
			}
		}

		
		long ans = 0;
		for (int i = 0 ; i < n ; i++) {
			if (rightBigger[i] != -1) {
				ans = Math.max(ans, a[i] ^ rightBigger[i]);
			}
			if (leftBigger[i] != -1) {
				ans = Math.max(ans, a[i] ^ leftBigger[i]);				
			}
		}
		out.println(ans);
		out.flush();
	}
	

	public static void debug(Object... o) {
		System.err.println(Arrays.deepToString(o));
	}
}
