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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Stack;
import java.util.StringTokenizer;

/**ID: sinbadc1
LANG: JAVA
TASK: wormhole
*/
public class Main{
    public static void main(String[] args)throws IOException{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int n=Integer.parseInt(br.readLine());
        int[] ar=new int[n];
        String s=br.readLine();
        StringTokenizer st=new StringTokenizer(s);
        for(int i=0;i<n;i++){
            ar[i]=Integer.parseInt(st.nextToken());
        }
        Stack<Integer> stack=new Stack<>();
        int res=-1;
        for(int i=0;i<n;i++){
            int curr=ar[i];
            if(stack.isEmpty()){
                stack.push(curr);
            }
            else{
                while(!stack.isEmpty() && curr>stack.peek()){
                    res=Math.max(res, curr^stack.pop());
                }
                if(!stack.isEmpty()){
                    res=Math.max(res, curr^stack.peek());
                }
                stack.push(curr);
            }
        }
        System.out.println(res);
    }
}
