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.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.Collections;
import java.util.Deque;
import java.util.Scanner;

import static java.lang.Math.cos;
import static java.lang.Math.sin;

public class D172 {
    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedInputStream(System.in));
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int n = in.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
        }
        Deque<Integer> stack = new ArrayDeque<Integer>();
        stack.add(a[0]);
        int ans = 0;
        for (int i = 1; i < n; i++) {
            while (stack.size() > 0 && stack.peek() < a[i]) {
                stack.pop();
            }
            if (stack.size() > 0) ans = Math.max(ans, stack.peek() ^ a[i]);
            stack.push(a[i]);
        }
        stack.clear();
        stack.add(a[n - 1]);
        for (int i = n - 2; i >= 0; i--) {
            while (stack.size() > 0 && stack.peek() < a[i]) {
                stack.pop();
            }
            if (stack.size() > 0) ans = Math.max(ans, stack.peek() ^ a[i]);
            stack.push(a[i]);
        }
        out.println(ans);
        out.close();
    }
}
