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}.
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
        Scanner sc = new Scanner(System.in);
        Stack<Integer> stk = new Stack<Integer>();
        int n = sc.nextInt();
        int a[] = new int[n];
        for(int i=0;i<n;i++)
        a[i] = sc.nextInt();
        int ans = 0;
        for(int i=0;i<n;i++)
        {
            
            while(!stk.empty())
            {
                if(a[i]<stk.peek())
                {
                    ans = (Math.max(ans,stk.peek()^a[i]));
                    break;
                }
                else
                {
                    ans = (Math.max(ans,stk.peek()^a[i]));
                    stk.pop();
                }
            }
            stk.push(a[i]);
        }
        System.out.println(ans);
	}
}
