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}.
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5;
int s[N][20], a[N], ans;
int get(int l, int r) {
  int len = r - l + 1, j = log2(len);
  len = 1 << j;
  return max(s[l][j], s[r - len + 1][j]);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    s[i][0] = a[i];
  }
  for (int j = 1; j < 20; j++) {
    int len = 1 << j;
    for (int i = 0; i + len - 1 < n; i++)
      s[i][j] = max(s[i][j - 1], s[i + len / 2][j - 1]);
  }
  for (int i = 0; i < n; i++) {
    int cur = i;
    for (int j = 20; j >= 0; j--) {
      int len = 1 << j;
      if (cur - len < 0) continue;
      if (s[cur - len][j] <= a[i]) cur -= len;
    }
    if (cur) ans = max(ans, a[i] ^ a[cur - 1]);
    cur = i;
    for (int j = 20; j >= 0; j--) {
      int len = 1 << j;
      if (cur + len >= n) continue;
      if (s[cur + 1][j] <= a[i]) cur += len;
    }
    if (cur + 1 < n) ans = max(ans, a[i] ^ a[cur + 1]);
  }
  printf("%d", ans);
  return 0;
}
