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 CF281D{
    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] s;
        StringBuffer sb = new StringBuffer();
        s = br.readLine().split(" ");
        int n = Integer.parseInt(s[0]);
        s = br.readLine().split(" ");
        int a[] = new int[n];
        Stack<Integer> st = new Stack<Integer>();
        int max = -1;

        for(int i=0;i<n;i++){
            a[i] = Integer.parseInt(s[i]);
        }
        st.push(a[0]);
        for(int i=1;i<n;i++){
            int top = st.peek();
            if(top > a[i]){
                st.push(a[i]);
                //System.out.println("Push:"+a[i]);
                max = Math.max(max, top ^ a[i]);
                continue;
            }

            while(!st.isEmpty() && top < a[i]){
                max = Math.max(max, top ^ a[i]);
                //System.out.println("Pop"+a[i]);
                st.pop();
                if(!st.isEmpty())
                top = st.peek();
            }
            if(!st.isEmpty()){
                top = st.peek();
                max = Math.max(max, top ^ a[i]);
            }
            st.push(a[i]);
            //System.out.println("Push"+a[i]);
            max = Math.max(max, top ^ a[i]);
        }
        System.out.println(max);
    }
}
