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;
import java.util.StringTokenizer;

public class Xor {

    private static long max = 0;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        br.readLine();
        solve(br.readLine());
    }

    private static void solve(String input) {
        max = 0;
        Stack<Long> stack = new Stack<>();
        StringTokenizer tokens = new StringTokenizer(input);
        int pre = Integer.MAX_VALUE;
        while (tokens.hasMoreTokens()) {
            int num = Integer.parseInt(tokens.nextToken());
            if (num > pre) {
                de(stack, num);
            } else en(stack, num);

            pre = num;
        }
        if (!stack.isEmpty()) setResult(stack);
        System.out.println(max);
    }

    private static void setResult(Stack<Long> stack) {
        long pre = stack.pop();
        while (!stack.isEmpty()) {
            long num = stack.pop();
            checkMax(pre ^ num);
            pre = num;
        }
    }

    private static void checkMax(long num) {
        if (num > max) max = num;
    }

    private static void en(Stack<Long> stack, long num) {
        stack.push(num);
    }

    private static void de(Stack<Long> stack, long num) {
        while (!stack.isEmpty() && stack.peek() < num) {
            long pop = stack.pop();
            checkMax(num ^ pop);
            if (!stack.isEmpty()) checkMax(stack.peek() ^ pop);
        }
        en(stack, num);
    }
}


















