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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Stack;
import java.util.StringTokenizer;

public class MaximumXorSecondary {

    public void solve() throws IOException {
        BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(rd.readLine());
        StringTokenizer tkn = new StringTokenizer(rd.readLine());

        Stack<Integer> st = new Stack<>();
        int maxXor = 0, val;
        for ( int i = 0; tkn.hasMoreTokens(); ++i ) {
            val = Integer.parseInt(tkn.nextToken());
            while ( !st.empty() ) {
                maxXor = Math.max(maxXor, st.peek() ^ val);
                if ( st.peek() > val )
                    break;
                st.pop();
            }
            st.push(val);
        }

        System.out.println(maxXor);
    }

    public static void main(String[] args) throws Exception {
        new MaximumXorSecondary().solve();
    }
}
