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.*;
public class MyClass {
    public static void main(String args[]) {
       Scanner sc=new Scanner(System.in);
       int n=sc.nextInt();
       int a[]=new int[n];
       for(int i=0;i<n;i++)
       a[i]=sc.nextInt();
       
       Stack s=new Stack();
       s.push(a[0]);
       long ans=0;
      long mx=-1;
       for(int i=1;i<n;i++)
       {
           if(a[i]<(int)(s.peek()))
           {
                    ans=a[i]^((int)s.peek());
                   // System.out.println("ashjasd "+ans);
               s.push(a[i]);
               if(ans>mx)
               mx=ans;
           }
           else
           {
               while((!s.empty()&&(int)s.peek()<a[i]))
               {
                   
                   ans=(int)s.pop()^(a[i]);
                  // System.out.println("akdas "+ans);
                   if(ans>mx)
                   mx=ans;
               }
               if(!s.empty())
               {
               ans=(int)s.peek()^(a[i]);
               if(ans>mx)
               mx=ans;}
               s.push(a[i]);
               
           }
           
       }
       
     
       System.out.println(mx);
           
       
    }
}