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.*;
import java.util.*;
public class Main {
    public static void main(String arg[]) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        String[] ip = br.readLine().split(" ");
        int[] arr = new int[ip.length];
        for(int i=0; i<arr.length; i++) arr[i] = Integer.parseInt(ip[i]);
        int max = -1;
        Stack<Integer> st = new Stack<>();
        for(int i=0; i<arr.length; i++) {
            while(!st.empty()) {
                int top = st.peek();
                max = max < (top ^ arr[i]) ? top ^ arr[i] : max;
                if(top > arr[i]) break;
                st.pop();
            }
            st.push(arr[i]);
        }
        
        System.out.println(max);
        
    }
}