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.util.*;
import java.lang.*;
import java.text.DecimalFormat;
import java.text.NumberFormat;
import java.io.*;
public class Solution {
	static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try{
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
	public int[] readIntArray(int n) {
			int[] arr = new int[n];
			for(int i=0; i<n; ++i)
				arr[i]=nextInt();
			return arr;
		}
	public long[] readLongArray(int n) {
		long[] arr = new long[n];
		for(int i=0; i<n; ++i)
			arr[i]=nextLong();
		return arr;
	}
        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try{
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
	static void solve(String query) {
		
	}
	public static void main(String args[]) {
        FastReader sc = new FastReader();
        long start = System.currentTimeMillis();
        	
        int n = sc.nextInt();
        
        int[] arr = sc.readIntArray(n);
        long max = 0;
        Stack<Integer> thestack = new Stack<>();
    	for(int i=0; i<n; ++i) {
    		while(!thestack.empty()&&arr[i]>=thestack.peek()) {
    			long a=thestack.peek();
    			thestack.pop();
    			max= Math.max(a^arr[i], max);
    		}
    		thestack.push(arr[i]);
    	}
    	thestack=new Stack<Integer>();
    	for(int i=n-1; i>=0; --i) {
    		while(!thestack.empty()&&arr[i]>=thestack.peek()) {
    			int a=thestack.peek();
    			thestack.pop();
    			max=Math.max(a^arr[i], max);
    		}
    		thestack.push(arr[i]);
    	}
    	System.out.println(max);
     
        	long end = System.currentTimeMillis();
        	NumberFormat formatter = new DecimalFormat("#0.00000");
        	//System.out.print("Execution time is " + formatter.format((end - start) / 1000d) + " seconds");
        }
	}
