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 double eps = 1.0e-11;
const double pi = acos(-1.0);
const int N = 100100;
int a[N];
int bit, n;
int Solve() {
  int res = 0;
  for (long long i = 0; i < n - 1; i++) {
    if ((a[i] & (1 << bit)) != 0 && (a[i + 1] & (1 << bit)) == 0) {
      int mx1 = max(a[i], a[i + 1]);
      int mx2 = min(a[i + 1], a[i]);
      res = max(res, mx1 ^ mx2);
      while (i + 2 < n && (a[i + 2] & (1 << bit)) == 0) {
        if (a[i + 2] > mx1) {
          mx2 = mx1;
          mx1 = a[i + 2];
        } else {
          if (a[i + 2] > mx2) {
            mx2 = a[i + 2];
          }
        }
        res = max(res, mx1 ^ mx2);
        ++i;
        if (i + 2 == n) return res;
      }
    }
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (long long i = 0; i < n; i++) scanf("%d", a + i);
  int mx = *max_element(a, a + n);
  bit = 0;
  while ((1 << bit) <= mx) ++bit;
  --bit;
  int res = 0;
  while (bit >= 0) {
    res = max(res, Solve());
    reverse(a, a + n);
    res = max(res, Solve());
    --bit;
  }
  cout << res << endl;
  return 0;
}
