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 MaxXORSecondary {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		Scanner in = new Scanner(System.in);
		
		int n=in.nextInt();
		
		int a[]=new int[n];
		
		for(int i=0;i<n;i++) {
			a[i]=in.nextInt();
		}
		
		Stack<Integer> s= new Stack<Integer>();
		
		
		int ans=-1;
		
		for(int i=0;i<n;i++)
		   {
		      while(!s.isEmpty())
		       {
		          if(s.peek()<a[i])
		           {
		            ans=Math.max(ans,s.peek() ^ a[i]);
		            s.pop();
		    }
		    else
		    {
		      ans=Math.max(ans,s.peek() ^ a[i]);
		      break;
		    }
		    }
		    s.push(a[i]);
		   }
		
		System.out.println(ans);

	}

}
