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 prepare{
   static int n,k;
   static StringBuilder ans; 
   static HashMap<Integer,Integer> map=new HashMap<>(); 
   public static void main(String[] args) throws IOException{ 
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); 
        ans=new StringBuilder(); 
        PrintWriter out=new PrintWriter(System.out); 
        int t=1;
        while(t-->0){
            n=Integer.parseInt(br.readLine().trim());
            StringTokenizer tok=new StringTokenizer(br.readLine()," ");
            int[] arr=new int[n];
            for(int i=0;i<n;i++){
                arr[i]=Integer.parseInt(tok.nextToken());
            }
            Stack<Integer> left=new Stack<>();
            long max=0;
            for(int i=0;i<n;i++){
                if(left.isEmpty()){
                    left.push(arr[i]);
                }
                else{
                    if(left.peek()>arr[i]){
                        max=Math.max(max, arr[i]^left.peek());
                        left.push(arr[i]);
                    }
                    else{
                        while(!left.isEmpty() && left.peek()<arr[i]){
                            max=Math.max(max, left.peek()^arr[i]);
                            left.pop();
                        }
                        if(!left.isEmpty())
                        max=Math.max(max, left.peek()^arr[i]);
                        left.push(arr[i]);
                    }
                }
            }     
            ans.append(max+"\n");
        }
        out.println(ans); 
        out.close(); 
    }
}