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.lang.Math;

public class maxxorsec{
	public static long [] array = new long[100010];
	public static int [] max_left = new int[100010];
	public static int [] max_right = new int[100010];
	public static void main(String[] args){
		Scanner input = new Scanner(System.in);
		int n = input.nextInt();
		for (int i=0;i<n;i++)
			array[i] = input.nextLong();
		max_left[0] = -1;
		for (int i=1;i<n;i++){
			int j = i-1;
			while(j!=-1 && array[j]<array[i])
				j = max_left[j];
			max_left[i] = j;
		}
		max_right[n-1] = -1;
		for (int i=n-2;i>=0;i--){
			int j=i+1;
			while(j!=-1 && array[j]<array[i])
				j = max_right[j];
			max_right[i] =j; 
		}
		long result = 0;
		for (int i=0;i<n;i++){
			if (max_right[i]!= -1 && ((array[i] ^ array[max_right[i]]) > result))
				result = array[i]^array[max_right[i]];
			if (max_left[i]!= -1 &&  ((array[i] ^ array[max_left[i]]) > result))
				result = array[i]^array[max_left[i]];
		}
		System.out.println(result);
	}
}
