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

public class B {

	int count(int[] a) {
		int res = 0;
		int maxlen = 0;
		int n = a.length;
		int[] type = new int[n];
		while (true) {
			int min = a[0];
			maxlen = 0;
			for (int i = 0; i < n; i++) {
				maxlen = Math
						.max(maxlen, Integer.toBinaryString(a[i]).length());
				min = Math.min(min, a[i]);
			}
			if (Integer.toBinaryString(min).length() != maxlen || maxlen == 1) {
				break;
			}
			for (int i = 0; i < n; i++) {
				a[i] = a[i] - (1 << (maxlen - 1));
			}
		}
		if (maxlen == 1) {
			int[] b = a.clone();
			Arrays.sort(b);
			return b[n - 1] - b[0];
		}

		for (int i = 0; i < n; i++) {
			if (maxlen == Integer.toBinaryString(a[i]).length()) {
				type[i] = 1;
			} else {
				type[i] = 0;
			}
		}
		for (int i = 1; i < n; i++) {
			if (type[i] == 1 && type[i - 1] == 0) {
				int sec = Integer.MIN_VALUE;
				for (int j = i - 1; (j >= 0) && (type[j] == 0); --j) {
					sec = Math.max(sec, a[j]);
					res = Math.max(res, sec ^ a[i]);
				}
			}
		}

		return res;
	}

	void count() throws IOException {
		int n = nextInt();
		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = nextInt();
		}
		int k = count(a);
		for (int i = 0; n - i - 1 > i; i++) {
			int t = a[i];
			a[i] = a[n - i - 1];
			a[n - i - 1] = t;
		}
		out.println(Math.max(k, count(a)));
	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer str;
	String readline;

	String nextToken() throws IOException {
		while (str == null || !str.hasMoreTokens()) {
			readline = in.readLine();
			if (readline == null) {
				return null;
			} else {
				str = new StringTokenizer(readline);
			}
		}
		return str.nextToken();
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	void run() throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		count();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new B().run();
	}
}