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.StringTokenizer;

public class MaximumXorSecondary {

	public static void main(String[] args) throws IOException {
		FastScanner sc = new FastScanner();
		int longitud = (int) sc.nextLong();
		long[] numeros = new long[longitud];
		for (int i = 0; i < numeros.length; i++) {
			numeros[i] = sc.nextLong();
		}

		long diferenciaPot2 = 1;

		for (int i = 0; i < numeros.length; i++) {
			if (i > 0) {
				// menor o menor o igual, esa es la cuestion
				while ((numeros[i] ^ numeros[i - 1]) >= diferenciaPot2) {
					diferenciaPot2 *= 2;
				}
			}
		}

		for (int i = 0; i < numeros.length; i++) {
			numeros[i] %= diferenciaPot2;
		}

		int anterior = -1, actual = 0, proximo = 1;
		while (numeros[actual] < diferenciaPot2 / 2) {
			actual++;
			proximo++;
		}

		while (proximo < longitud && numeros[proximo] < diferenciaPot2 / 2) {
			proximo++;
		}

		long maximoXor = 0;
		while (actual < longitud) {
			long sgdMax = 0;
			for (int i = actual - 1; i > anterior; i--) {
				if (numeros[i] >= sgdMax) {
					sgdMax = numeros[i];
					maximoXor =
							Math.max(maximoXor, numeros[i] ^ numeros[actual]);
				}
			}
			sgdMax = 0;
			for (int i = actual + 1; i < proximo; i++) {
				if (numeros[i] >= sgdMax) {
					sgdMax = numeros[i];
					maximoXor =
							Math.max(maximoXor, numeros[i] ^ numeros[actual]);
				}
			}
			anterior = actual;
			actual = proximo;
			do {
				proximo++;
			} while (proximo < longitud
					&& numeros[proximo] < diferenciaPot2 / 2);

		}
		System.out.println(maximoXor);
	}
}

class FastScanner {

	private BufferedReader br;
	private StringTokenizer st;

	public FastScanner() {
		br = new BufferedReader(new InputStreamReader(System.in));
	}

	public long nextLong() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			st = new StringTokenizer(br.readLine());
		}
		return Long.parseLong(st.nextToken());
	}

}
