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 + 1;
int arr[N];
int nxt[N];
int ans;
int n;
void input() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
  }
}
int calc() {
  stack<int> s;
  fill(nxt, nxt + N, -1);
  for (int i = 0; i < n; ++i) {
    while (!s.empty() && arr[i] >= arr[s.top()]) {
      nxt[s.top()] = i;
      s.pop();
    }
    s.push(i);
  }
  int res = 0;
  for (int i = 0; i < n; ++i) {
    if (nxt[i] != -1) {
      res = max(res, arr[i] ^ arr[nxt[i]]);
    }
  }
  return res;
}
void solve() {
  ans = max(ans, calc());
  reverse(arr, arr + n);
  ans = max(ans, calc());
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  input();
  solve();
}
