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;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.Stack;
import java.io.InputStream;
import java.util.Vector;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author dy
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskD {
    public void solve(int testNumber, Scanner in, PrintWriter out) {//tle
        int N = in.nextInt();
        int[] d = new int[N];
        for(int i = 0; i < N; ++i) d[i] = in.nextInt();
        Stack<Integer> stack = new Stack<Integer>();
        stack.push(d[0]);
        int ans = 0;
        for(int i = 1; i < N; ++i){
            while(!stack.empty() && stack.peek() < d[i])
                stack.pop();
            if(!stack.empty())
                ans = Math.max(d[i] ^ stack.peek(), ans);
            stack.push(d[i]);
        }
        stack.clear();
        stack.push(d[N - 1]);
        for(int i = N - 2; i >= 0; --i){
            while(!stack.empty() && stack.peek() < d[i])
                stack.pop();
            if(!stack.empty())
                ans = Math.max(d[i] ^ stack.peek(), ans);
            stack.push(d[i]);
        }
        out.println(ans);
    }
}
