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 class MXSEC {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		int n = Integer.parseInt(br.readLine());
		
		String[] ip = br.readLine().split(" ");
		long[] ary = new long[n];
		
		for(int i = 0 ; i < n  ; i++) {
			ary[i] = Long.parseLong(ip[i]);
		}
		
		System.out.println(solve(ary));
	}
	
	public static long solve (long[] ary) {
		Deque<Long> stack = new ArrayDeque<>();
		
		long max_xor = Long.MIN_VALUE;
		
		int i = 0;
		int n = ary.length; 
		
		while ( i < n ) {
			if(stack.isEmpty() || stack.peek() > ary[i]) {
				if(!stack.isEmpty()) {
				long temp = stack.peek() ^ ary[i];
				if ( max_xor < temp )
					max_xor = temp;
				}
				stack.push(ary[i++]);
			}
			else {
				long temp = stack.pop() ^ ary[i];
				if ( max_xor < temp )
					max_xor = temp;
			}
		}
		
		/*while(!stack.isEmpty()) {
			long a=0;
			long b=0;
			
			a = stack.pop();
			
			if(stack.isEmpty())
				break;
			
			b = stack.peek();
			long temp = a^b;
			if ( max_xor < temp )
				max_xor = temp;
			
		}*/
		
		return max_xor;
	}
}
