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.*;
import java.io.*;

public class Main{

    BufferedReader in;
    StringTokenizer str = null;
    PrintWriter out;

    private String next() throws Exception{
    	if (str == null || !str.hasMoreElements())
    	    str = new StringTokenizer(in.readLine());
    	return str.nextToken();
    }
    
    private int nextInt() throws Exception{
	   return Integer.parseInt(next());
    }
    
    public void run() throws Exception{
    	in = new BufferedReader(new InputStreamReader(System.in));
    	out = new PrintWriter(System.out);

        int n = nextInt();
        int stack[] = new int[n];
        int yk = -1;
        int max = 0;
        for(int i=0;i<n;i++){
            int x = nextInt();
            while(yk >= 0 && x > stack[yk]){
                max = Math.max(max, x ^ stack[yk]);
                yk--;
            }
            stack[++yk] = x;
            if (yk > 0)
                max = Math.max(max, x ^ stack[yk-1]);
        }	
        out.println(max);
        out.close();
    }

    public static void main(String args[]) throws Exception{
	new Main().run();
    }
}