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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Stack;

public class Main {
    static String[] data;

    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(reader.readLine());
        data = reader.readLine().split(" ");
        int[] input = new int[n];
        for (int i = 0; i < n; i++) {
            input[i] = Integer.parseInt(data[i]);
        }
        Stack<Integer> s = new Stack<>();
        s.push(input[0]);
        int maxLuckyNum = 0;
        for (int i = 1; i < n; i++) {
            while (!s.isEmpty() && s.peek() < input[i]) {
                maxLuckyNum = Math.max(maxLuckyNum, s.pop() ^ input[i]);
            }
            if(!s.isEmpty()){
                maxLuckyNum = Math.max(maxLuckyNum,s.peek() ^ input[i]);
            }
            s.push(input[i]);
        }
        int temp1 = s.pop();
        while (!s.isEmpty()) {
            int temp2 = s.pop();
            maxLuckyNum = Math.max(maxLuckyNum, temp1 ^ temp2);
            temp1 = temp2;
        }
        System.out.println(maxLuckyNum);
    }
}