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 practice;



import java.util.*;


public class Main
{
   
    
    public static void main (String[] args) 
    {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int[] arr = new int[n];
        for(int i=0; i<n; i++)
        {
            arr[i] = scan.nextInt();
        }
        long max = -1;
        int top;
        Stack<Integer> stack = new Stack<>();
        int i=0;
        while(i<n)
        {
            int a = arr[i];
            if(stack.isEmpty())
            {
                stack.push(a);
                i++;
            }
            
            else
            {
                top = stack.peek();
                if(stack.peek()>a)
                {
                    stack.push(arr[i]);
                    i++;
                }
            
                else
                    stack.pop();
            
                max = Math.max(max, top^a);
            }
            
        }
        System.out.println(max);
                
        
        
    }
}