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.io.*;

public class Main implements Runnable {
        int answer;
	public void solve() throws IOException {
		int N = nextInt();
                int[] a = new int[N];
                for(int i = 0; i < a.length; i++) a[i] = nextInt();
                
                //arrange in LIS and see the magic
                answer = 0;
                doit(a);
                
                //reverse
                for(int i = 0; i < N / 2; i++){
                    int t = a[i];
                    a[i] = a[N - 1 - i];
                    a[N - 1 - i] = t;
                }
                
                doit(a);
                
                System.out.println(answer);
	}
        
        private void doit(int[] a){
            Stack<Integer> stk = new Stack<>();
            for(int i = 0; i < a.length; i++){
                while(!stk.isEmpty() && a[stk.peek()] < a[i]) stk.pop();
                if(!stk.isEmpty()) answer = Math.max(answer, a[stk.peek()] ^ a[i]);
                stk.push(i);
            }
        }

	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	//-----------------------------------------------------------
	public static void main(String[] args) {
		new Main().run();
	}

        public void debug(Object... arr){
            System.out.println(Arrays.deepToString(arr));
        }

        public void print1Int(int[] a){
                for(int i = 0; i < a.length; i++)
                        System.out.print(a[i] + " ");
                System.out.println();
        }
        
        public void print2Int(int[][] a){
                for(int i = 0; i < a.length; i++){
                        for(int j = 0; j < a[0].length; j++){
                                System.out.print(a[i][j] + " ");
                        }
                        System.out.println();
                }
        }
        
	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			tok = null;
			solve();
			in.close();
		} catch (IOException e) {
			System.exit(0);
		}
	}

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

	BufferedReader in;
	StringTokenizer tok;
}