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.BufferedWriter;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.Stack;

/* Name of the class has to be "Main" only if the class is public. */
public class Solution {
	public static void main(String[] args) throws java.lang.Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
		String inp[];
		int n = Integer.parseInt(br.readLine());
		int arr[] = new int[n];
		inp = br.readLine().split(" ");
		for (int i = 0; i < n; i++)
			arr[i] = Integer.parseInt(inp[i]);
		Stack<Integer> st = new Stack<>();
		st.push(0);
		long ans = arr[0] ^ arr[1];
		for (int i = 1; i < n; i++) {
			while (!st.isEmpty() && arr[st.peek()] <= arr[i])
				st.pop();
			if (!st.isEmpty())
				ans = Math.max(ans, arr[i] ^ arr[st.peek()]);
			st.push(i);
		}
		st.clear();
		st.push(0);
		for (int i = 1; i < n; i++) {
			while (!st.isEmpty() && arr[st.peek()] < arr[i])
				ans = Math.max(ans, arr[i] ^ arr[st.pop()]);
			st.push(i);
		}
		bw.write(String.valueOf(ans) + "\n");
		bw.flush();
	}
}
