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;
int vet[100000], n, res;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &vet[i]);
  set<pair<int, int> > s;
  s.insert(make_pair(vet[n - 1], n - 1));
  for (int i = n - 2; i >= 0; --i) {
    int now = vet[i];
    pair<int, int> p = make_pair(0, 0);
    bool g = 0;
    while (1) {
      typeof(s.begin()) it = s.lower_bound(p);
      if (it == s.end()) break;
      p = *it;
      if (g && p.first > now) break;
      if (p.first > now) g = 1;
      res = max(res, now ^ p.first);
      p.first++;
    }
    typeof(s.begin()) it = s.lower_bound(make_pair(now, i));
    if (it != s.begin()) {
      --it;
      while (1) {
        bool b = it == s.begin();
        if (it->second > i)
          s.erase(it--);
        else
          break;
        if (b) break;
      }
    }
    s.insert(make_pair(now, i));
  }
  printf("%d\n", res);
}
