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}.
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF280B extends PrintWriter {
	CF280B() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF280B o = new CF280B(); o.main(); o.flush();
	}

	int[] aa, stack;
	int solve(int n) {
		int cnt = 0, ans = 0;
		for (int i = 0; i < n; i++) {
			int a = aa[i], b;
			while (cnt > 0 && (b = stack[cnt - 1]) < a) {
				ans = Math.max(ans, a ^ b);
				cnt--;
			}
			stack[cnt++] = a;
		}
		return ans;
	}
	void main() {
		int n = sc.nextInt();
		aa = new int[n];
		for (int i = 0; i < n; i++)
			aa[i] = sc.nextInt();
		stack = new int[n];
		int ans = solve(n);
		for (int i = 0, j = n - 1; i < j; i++, j--) {
			int tmp = aa[i]; aa[i] = aa[j]; aa[j] = tmp;
		}
		ans = Math.max(ans, solve(n));
		println(ans);
	}
}
