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.StringTokenizer;
import static java.lang.Math.*;

public class MaximumXorSecondary {

	public static void main(String[] args) throws IOException{
		BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(r.readLine());
		StringTokenizer str = new StringTokenizer(r.readLine());
		int[] arr = new int[n];
		for(int i = 0; i < n; i++){
			arr[i] = Integer.parseInt(str.nextToken());
		}
		int[] aux = new int[n];
		aux[0] = arr[0];
		int max = 0;
		int p = 0;
		for(int i = 1; i < n; i++){
			while(p >= 0 && aux[p] < arr[i]){
				max = max(max, arr[i]^aux[p]);
				p--;
			}
			if(p >= 0){
				max = max(max, arr[i]^aux[p]);
			}
			aux[++p] = arr[i];
		}
		System.out.println(max);
	}

}
