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.*;

public class test {
    int INF = (int)1e9;
    int MOD = 1000000007;
    
    
    void solve(InputReader in, PrintWriter out) throws IOException {
        int n = in.nextInt();
        int max = 0;
        Stack<Integer> stack = new Stack<>();
        for(int i=0; i<n; i++) {
            int a = in.nextInt();
            while(stack.size()>0) {
                if(stack.peek() <= a) {
                    int x = stack.pop();
                    max = Math.max(max, x^a);
                } else {
                    break;
                }
            }
            if(stack.size()>0) {
                int x = stack.peek();
                max = Math.max(max, x^a);
            }
            stack.push(a);
        }
//        int max = 0;
//        int a = stack.pop();
//        while(stack.size()>0) {
//            int b = stack.pop();
//            max = Math.max(max, a^b);
//            a = b;
//        }
        out.println(max);
    }
    
    
    public static void main(String[] args) throws IOException {
        InputReader in = new InputReader();
        PrintWriter out = new PrintWriter(System.out);
        int t = 1;//in.nextInt();
        while(t-- >0) {
            new test().solve(in, out);
        }
        out.close();
    }
    
    static class InputReader {
        static BufferedReader br;
        static StringTokenizer st;
    
        public InputReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        
        String next() {
            while (st == null || !st.hasMoreTokens()) {
                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());
        }
        
        double nextDouble() {
            return Double.parseDouble(next());
        }
        
    }
    
    class Pair implements Comparable<Pair> {
        long f, s;
        
        Pair(long f, long s) {
            this.f=f;
            this.s=s;
        }
        
        public int hashCode() {
            int hf = (int) (f ^ (f >>> 32));
            int hs = (int) (s ^ (s >>> 32));
            return 31 * hf + hs;
        }
        
        public boolean equals(Object o) {
            Pair other = (Pair) o;
            return s == other.s && f == other.f;
        }
        
        public int compareTo(Pair p) {
            return Long.compare(this.s, p.s);
        }
        
        public String toString() {
            return "(" + f + ", " + s + ")";
        }
    }
    
}