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.PrintStream;
import java.util.Scanner;
import java.util.Stack;

public class Main {
    private PrintStream out;
    private Scanner in;

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        new Main(in, System.out).solve();
        in.close();
    }

    public Main(Scanner in, PrintStream out) {
        this.in = in;
        this.out = out;
    }

    public void solve() {
        // TODO: Solve the problem
        int n = in.nextInt();
        int[] nums = new int[n];
        int i = 0;

        for(; i<n; ++i) {
            nums[i] = in.nextInt();
        }

        Stack<Integer> s = new Stack<>();
        i = 0;
        int max = 0;
        while( i<n ) {
            if( s.empty() || s.peek() > nums[i] ) {
                if( !s.empty() ) {
                    int no = s.peek() ^ nums[i];
                    if( no > max ) max = no;
//                    System.out.println("max=" + max + ", first=" + nums[i] + ", second=" + s.peek() + ", ans=" + no + " stack=" + s);
                }
                s.push(nums[i++]);
            } else {
                int top = s.pop();
                int no = top ^ nums[i];
                if ( no > max ) max = no;
//                System.out.println("max=" + max + ", first=" + top + ", second=" + nums[i] + ", ans=" + no + " stack=" + s);
            }
        }

        while( !s.empty() ) {
            int top = s.pop();
            if( !s.empty() ) {
                int no = s.peek() ^ top;
                if( no > max ) max = no;
//                System.out.println("max=" + max + ", first=" + top + ", second=" + s.peek() + ", ans=" + no + " stack=" + s);
            }
        }

        out.println(max);
    }


}
