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.*;

public class D {
	static int solve(int [] a){
		int res = 0;
		Stack<Integer> stack = new Stack<Integer>();
		for (int i : a){
			while (!stack.isEmpty() && stack.peek().compareTo(i) < 0){
				stack.pop();
			}
			if (!stack.isEmpty()){
				res = Math.max(res, stack.peek()^i);
			}
			stack.push(i);
		}
		return res;
	}
	public static void main(String [] args){
		try (Scanner s = new Scanner(System.in)){
			final int n = s.nextInt();
			final int [] a = new int [n];
			final int [] b = new int [n];
			for (int i = 0; i < n; ++i){
				b[n - 1 - i] = a[i] = s.nextInt();
			}
			System.out.println(Math.max(solve(a), solve(b)));
		}
	}
}
