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.util.Scanner;

/**
 * Created by Dudin Victor (duviteck@gmail.com)
 * SPbSU, Faculty of Mathematics and Mechanics
 * 2013 year
 */
public class Problem_D_MaxXOR {

    private static int[] stack;
    private static int stackTop = -1;
    private static int bestXOR = 0;

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();

        stack = new int[n];
        for (int i = 0; i < n; i++) {
            int num = s.nextInt();
            stackPush(num);
        }

        while (stackTop > 0) {
            int xor = stack[stackTop] ^ stack[stackTop - 1];
            if (xor > bestXOR) {
                bestXOR = xor;
            }
            stackTop--;
        }

        System.out.println(bestXOR);
    }

    private static void stackPush(int num) {
        if (stackTop == -1) {
            stack[++stackTop] = num;
            return;
        }

        while (stackTop >= 0 && stack[stackTop] < num) {
            int xor = stack[stackTop] ^ num;
            if (xor > bestXOR) {
                bestXOR = xor;
            }
            stackTop--;
        }

        stack[++stackTop] = num;
        if (stackTop > 0) {
            int xor = stack[stackTop] ^ stack[stackTop - 1];
            if (xor > bestXOR) {
                bestXOR = xor;
            }
        }
    }
}
