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;
void accept(int arr[], long long size) {
  for (long long h = 0; h < size; h++) {
    cin >> arr[h];
  }
}
void accept_vector(vector<long long>& arr, long long n) {
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
}
void showlist(list<long long> ans) {
  for (auto it = ans.begin(); it != ans.end(); it++) cout << *it << " ";
}
void showstack(stack<long long> s) {
  if (s.empty()) return;
  int x = s.top();
  s.pop();
  showstack(s);
  cout << x << " ";
  s.push(x);
}
int main() {
  long long T;
  for (long long test_case = 0; test_case < 1; test_case++) {
    long long n, ans = 0;
    scanf("%lld", &n);
    vector<long long> a(n);
    stack<long long> st;
    accept_vector(a, n);
    for (long long i = 0; i < n; i++) {
      while (!st.empty()) {
        if (a[i] > st.top()) {
          ans = max(ans, st.top() ^ a[i]);
          st.pop();
        } else {
          ans = max(ans, st.top() ^ a[i]);
          break;
        }
      }
      st.push(a[i]);
    }
    cout << ans << endl;
  }
}
