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 java.math.*;
import java.awt.geom.*;

import static java.lang.Math.*;

public class Solution implements Runnable {
	
	public void solve () throws Exception {
		int n = sc.nextInt();
		int a [] = new int [n];
		for (int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
		}
		int result = 0;
		for (int bit = 30; bit >= 0; bit--) {
			int oneCount = 0;
			for (int i = 0; i < n; i++) {
				if (((1 << bit) & a[i]) > 0)
					oneCount++;
			}
			if (oneCount == 0 || oneCount == n) continue;
			for (int i = 0; i < n; i++) {
				if (((1 << bit) & a[i]) > 0) {
					int maxi = 0;
					for (int j = i + 1; j < n; j++) {
						if ((a[j] & (1 << bit)) > 0) break;
						maxi = max (maxi, a[j]);
						result = max (result, a[i] ^ maxi);
					}
				}
			}
			for (int i = n - 1; i >= 0; i--) {
				if (((1 << bit) & a[i]) > 0) {
					int maxi = 0;
					for (int j = i - 1; j >= 0; j--) {
						if ((a[j] & (1 << bit)) > 0) break;
						maxi = max (maxi, a[j]);
						result = max (result, a[i] ^ maxi);
					}
				}
			}
			break;
		}
		out.println(result);
	} 
	
	BufferedReader in;
	PrintWriter out;
	FastScanner sc;

	static Throwable uncaught;

	@Override
	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			sc = new FastScanner(in);
			solve();
		} catch (Throwable t) {
			Solution.uncaught = t;
		} finally {
			out.close();
		}
	}

	public static void main(String[] args) throws Throwable {
		Thread t = new Thread(null, new Solution(), "", (1 << 26));
		t.start();
		t.join();
		if (uncaught != null) {
			throw uncaught;
		}
	}

}

class FastScanner {

	BufferedReader reader;
	StringTokenizer strTok;

	public FastScanner(BufferedReader reader) {
		this.reader = reader;
	}

	public String nextToken() throws IOException {
		while (strTok == null || !strTok.hasMoreTokens()) {
			strTok = new StringTokenizer(reader.readLine());
		}
		return strTok.nextToken();
	}

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

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

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

}