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.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.Stack;


public class D {
	static PrintWriter out;
	static StreamTokenizer st;
	static BufferedReader bf;
	public static void main(String[] args) throws IOException {
		bf = new BufferedReader(new InputStreamReader(System.in));
		st = new StreamTokenizer(bf);
		out = new PrintWriter(new OutputStreamWriter(System.out));
		int  n = nextInt();
		Stack<Long> st = new Stack<Long>();
		long max = 0;
		for (int i = 0; i < n; i++) {
			long x = nextInt();
			while (!st.isEmpty() && x > st.peek())
				max = Math.max(max, x^st.pop());
			if (st.size() >= 1) max = Math.max(max, x^st.peek());
			st.push(x);
		}
		System.out.println(max);
	}
	private static String nextLine() throws IOException {
		return bf.readLine().trim();
	}
	private static double nextDouble() throws IOException{
		st.nextToken();
		return st.nval;
	}
	private static long nextLong()  throws IOException{
		st.nextToken();
		return (long) st.nval;
	}
	private static int nextInt() throws IOException {
		st.nextToken();
		return (int) st.nval;
	}

}
