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 MaximumXor{
  public static void main(String[] arg) throws IOException{
    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    int n= Integer.parseInt(br.readLine());
    String[] s=br.readLine().trim().split(" ");
    int[] A=new int[n];
    for(int i=0;i<n;i++)
        A[i]=Integer.parseInt(s[i]);
    //Arrays.sort(A);
    //System.out.println(Arrays.toString(A));
    Stack<Integer> st=new Stack<Integer>();
    long ans=0;
    for(int i=0;i<n;i++){
	   while(!st.isEmpty()){
	     if(st.peek()<A[i]){
		   ans=Math.max(ans,st.peek()^A[i]);
		   st.pop();
		 }
		 else{
			 ans=Math.max(ans,st.peek()^A[i]);
			 break;
		 }
	   }
	   st.push(A[i]);
       //System.out.println(st);
	}
	System.out.println(ans);
  }

}