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 static java.lang.System.*;
import java.util.*;
import static java.lang.Math.*;

public class D281{
    public static Scanner sc = new Scanner(in);
    public static Random rand=new Random();

    int count(int []a){
        Stack<Integer> stack = new Stack<Integer>();
        int count=  0;
        for (int i = 0; i<a.length; i++){
            while (!stack.isEmpty() && stack.peek() < a[i])
                stack.pop();
            if (!stack.isEmpty())
                count = max(count, stack.peek() ^ a[i]);
            stack.push(a[i]);
        }
        return count;
    }

    public void run(){
    	int n=sc.nextInt();
    	int[] a=nextIntArray(n);

    	int[] rev=new int[n];
    	for (int i = 0; i<n; i++)
    		rev[i] = a[n - i - 1];

        ln(max(count(a), count(rev)));// > and <
    }

    public static void main(String[] _) {
        new D281().run();
    }

    public int[] nextIntArray(int n){
        int[] res=new int[n];
        for(int i=0;i<n;i++){
            res[i]=sc.nextInt();
        }
        return res;
    }
    public static void pr(Object o) {
        out.print(o);
    }
    public static void ln(Object o) {
        out.println(o);
    }
    public static void ln() {
        out.println();
    }

} 