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.*;
public final class Maximum_Xor_Secondary
{
	public static void main(String[] args)
		{
			Scanner sc=new Scanner(System.in);
			int t=sc.nextInt();
			int arr[]=new int[t];
			for(int i=0;i<t;i++)
				arr[i]=sc.nextInt();
			Stack<Integer> st=new Stack<Integer>();
			int max_xor=0;
			for(int i=0;i<t;i++)
			{
				while(!st.empty())
				{
					if(arr[i]>st.peek())
					{
						max_xor=Math.max(max_xor,st.peek()^arr[i]);
						st.pop();
					}
					else
					{
						max_xor=Math.max(max_xor,st.peek()^arr[i]);
						break;
					}
				}
				st.push(arr[i]);
			}
			System.out.print(max_xor);
		}
}