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.io.*;
import java.util.*;
public final class Maxxor
{
	static class Input
	{
		BufferedReader br;
		StringTokenizer st;
		public Input()
		{
			br=new BufferedReader(new InputStreamReader(System.in)); 
		}
		String next()
		{
			while(st==null||!st.hasMoreElements())
			{
				try 
				{
					st=new StringTokenizer(br.readLine());
				}
				catch(IOException e)
				{
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		int nextInt()
		{
			return Integer.parseInt(next());
		}
	}
	public static void main(String[] args)
	{
		Input in=new Input();
		int n=in.nextInt();
		int xor=0;
		Stack<Integer> st=new Stack<Integer>();
		while(n-->0)
		{
			int tmp=in.nextInt();
			while(!st.isEmpty()&&tmp>st.peek())
			{
				int txor=st.peek()^tmp;
				if(txor>xor)
				{
					xor=txor;
				}
				st.pop();
			}
			if(!st.isEmpty()&&tmp<st.peek())
			{
				int txor=st.peek()^tmp;
				if(txor>xor)
				{
					xor=txor;
				}
			}
				st.push(tmp);
		}
		System.out.println(xor);
	}
}