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.*;
import java.io.*;

public class Main
{
	public static void main (String[] args) throws java.lang.Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		int[] arr = new int[n];
		String[] brr = br.readLine().split(" ");
		for(int i = 0; i < n; i++) {
			arr[i] = Integer.parseInt(brr[i]);
		}

		Stack<Integer> stk = new Stack<Integer>();
		stk.push(arr[0]);
		long max = 0;

		for(int i = 1; i < arr.length; i++) {
			if(stk.isEmpty() || stk.peek() > arr[i]) {
				max = Math.max(max, arr[i] ^ stk.peek());
				stk.push(arr[i]);
			} else {
				while(!stk.isEmpty() && stk.peek() < arr[i]) {
					max = Math.max(max, arr[i] ^ stk.pop());
				}
				if(!stk.isEmpty()) {
					max = Math.max(max, arr[i] ^ stk.peek());
				}
				stk.push(arr[i]);
			}
		}

		System.out.println(max);
	}
}