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.*;

public class B280 {
	public static void main(String[] args) throws FileNotFoundException {
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int n = in.nextInt();
		long a[] = new long[n];
		for (int i = 0; i < n; i++) {
			a[i] = in.nextInt();
		}
		long ans = 0;
		int last[] = new int[n];
		last[0] = -1;
		for (int i = 1; i < n; i++) {
			if (a[i] < a[i - 1]) {
				ans = Math.max(ans, a[i] ^ a[i - 1]);
				last[i] = i - 1;
			} else {
				int cur = i - 1;
				while ((a[cur] < a[i]) && (last[cur] != -1)) {
					ans = Math.max(ans, a[i] ^ a[cur]);
					cur = last[cur];
				}
				ans = Math.max(ans, a[i] ^ a[cur]);
				if (a[cur] > a[i]) {
					last[i] = cur;
				}
				else{
					last[i] = -1;
				}
			}
		}
		out.print(ans);
		out.close();
	}
}
