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 B {
    public static void main(String[] args) throws Exception {
        new B().solve();
        // new FileInputStream(new File("input.txt")),
        // new PrintStream(new FileOutputStream(new File("output.txt"))));
    }

    void solve() throws IOException {
        BufferedReader in = new BufferedReader(new
                InputStreamReader(System.in));
        Scanner sc = new Scanner(System.in);
        String[] sp;

        int n = Integer.parseInt(in.readLine());
        int[] a = new int[n];
        sp = in.readLine().split(" ");
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(sp[i]);
        }

        //
        int h;
        while (true) {
            int max = 0;
            int and = -1;
            for (int i = 0; i < n; i++) {
                max = Math.max(max, a[i]);
                and &= a[i];
            }
            if ((h = Integer.highestOneBit(max)) == Integer.highestOneBit(and)) {
                for (int i = 0; i < n; i++) {
                    a[i] -= h;
                }
            } else {
                break;
            }
        }

        //
        boolean[] one = new boolean[n];
        for (int i = 0; i < n; i++) {
            one[i] = (a[i] & h) != 0;
        }

        //
        int max = -1;
        for (int i = 0; i < n; i++) {
            if (one[i]) {

                int smax = -1;
                for (int jj = i - 1; jj >= 0 && !one[jj]; jj--) {
                    if (smax < a[jj]) {
                        smax = a[jj];
                        max = Math.max(max, a[i] ^ a[jj]);
                    }
                }
                smax = -1;
                for (int jj = i + 1; jj < n && !one[jj]; jj++) {
                    if (smax < a[jj]) {
                        smax = a[jj];
                        max = Math.max(max, a[i] ^ a[jj]);
                    }
                }
            }
        }
        System.out.println(max);
    }
}

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