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 mxors
{
	public static void main(String[] args)
	{
		Scanner sc=new Scanner(System.in);
		int t,i,j;
			int n=sc.nextInt();
			int max=0,smax=0,mxor=0;
			int a[]=new int[n];
			for(i=0;i<n;i++)
				a[i]=sc.nextInt();
			Stack<Integer> p=new Stack<Integer>();
			  p.push(a[0]);
			for(i=1;i<n;i++)
			{
				 if(a[i]<p.peek())
				{
					if((a[i]^p.peek())>mxor)
						mxor=a[i]^p.peek();
					p.push(a[i]);
				}
				else
				{
					while(p.peek()<a[i])
					{
						int y=p.pop();
						if((y^a[i])>mxor)
							mxor=y^a[i];
						if(p.isEmpty())
							break;
					}
					if(!p.isEmpty())
					{
						if((a[i]^p.peek())>mxor)
							mxor=a[i]^p.peek();
					}
					p.push(a[i]);
				}
			}
			smax=p.pop();
			if(!p.isEmpty())	
			{
			if((smax^p.peek())>mxor)
				mxor=smax^p.peek();
			}
			System.out.println(mxor);
		}
	}

