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 N;
int a[100010];
vector<pair<int, int> > p;
set<int> s;
int ans;
void func(int y) {
  int i;
  __typeof(s.begin()) itr = s.lower_bound(y);
  int z = (*itr);
  itr--;
  int x = (*itr);
  if (x != -1) ans = max(ans, (a[x] ^ a[y]));
  if (z != N) ans = max(ans, (a[y] ^ a[z]));
  s.insert(y);
}
int main(void) {
  int i;
  cin >> N;
  for ((i) = 0; (i) < (int)(N); (i)++) scanf("%d", &a[i]);
  for ((i) = 0; (i) < (int)(N); (i)++) p.push_back(make_pair(-a[i], i));
  sort(p.begin(), p.end());
  s.insert(-1);
  s.insert(N);
  for ((i) = 0; (i) < (int)(N); (i)++) func(p[i].second);
  cout << ans << endl;
  return 0;
}
