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