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

import static java.lang.Math.*;

public class Solution implements Runnable {

    class Element implements Comparable<Element> {
        
        int val, pos;
        
        public Element(int val, int pos) {
            this.val = val;
            this.pos = pos;
        }
        
        @Override
        public int compareTo(Element o) {
            return this.val - o.val;
        }
        
    }
    
    public void solve() throws Exception {
        int n = sc.nextInt();
        int[] s = new int[n];
        Element[] sorted = new Element[n];
        for (int i = 0; i < n; i++) {
            s[i] = sc.nextInt();
            sorted[i] = new Element(s[i], i);
        }
        
        Arrays.sort(sorted);
        
        TreeSet<Integer> set = new TreeSet<Integer>();
        
        int ans = 0;
        
        for (int i = n - 1; i >= 0; i--) {
            Integer ceil = set.ceiling(sorted[i].pos);
            if (ceil != null) {
                ans = max(ans, sorted[i].val ^ s[ceil]);
            }
            Integer floor = set.floor(sorted[i].pos);
            if (floor != null) {
                ans = max(ans, sorted[i].val ^ s[floor]);
            }
            set.add(sorted[i].pos);
        }
        
        out.println(ans);
    }
    
    static Throwable uncaught;
    
    BufferedReader in;
    FastScanner sc;
    PrintWriter out;
    
    @Override
    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
            sc = new FastScanner(in);
            solve();
        } catch (Throwable uncaught) {
            Solution.uncaught = uncaught;
        } finally {
            out.close();
        }
    }
    
    public static void main(String[] args) throws Throwable {
        Thread thread = new Thread (null, new Solution(), "", (1 << 26));
        thread.start();
        thread.join();
        if (Solution.uncaught != null) {
            throw Solution.uncaught;
        }
    }

}

class FastScanner {
    
    BufferedReader in;
    StringTokenizer st;
    
    public FastScanner(BufferedReader in) {
        this.in = in;
    }
    
    public String nextToken() throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
    
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }
    
    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }
    
    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
    
}