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.PrintWriter; //PrintWriter

public class R172_Div2_D
{
	public static void main(String[] args) {
		Scanner      in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		solve(in, out);
		out.close();
		in.close();	
	}
	
	public static void solve(Scanner in, PrintWriter out)
	{
		int n = in.nextInt();		
		int[] a = new int[n];
		for (int i = 0; i < n; i++) 
			a[i] = in.nextInt();
		
		int[] stack = new int[n];
        int ret = 0;
        //Find all pairs of numbers having smaller number(s) between,
        //and compute xor of the 2 numbers, and take maximum of all.
        //1. For a[i], find first number to its left that is bigger
        //2. For a[i] reversed, find first number to its left that is bigger
        for (int k = 1; k <= 2; k++)
        {
            if (k == 2) //Array.Reverse(a);
            {
            	for (int j = 0; j < n/2; j++)
            	{
            		int temp = a[j]; a[j] = a[n-j-1]; a[n-j-1] = temp;
            	}
            }

            int j = 0;
            for (int i = 0; i < n; i++)
            {
                while (j > 0 && stack[j - 1] < a[i]) j--;
                if (j > 0)
                    ret = Math.max(ret, stack[j - 1] ^ a[i]);
                stack[j++] = a[i];
            }
        }
		
		out.println(ret);		
	}
}
