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 hi() {
  static long long i = 0;
  cout << "Check Point : " << ++i << "\n";
  return;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  t = 1;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> v(n);
    for (long long i = 0; i < n; i++) {
      cin >> v[i];
    }
    stack<long long> right_next_greater_stk;
    stack<long long> left_next_greater_stk;
    unordered_map<long long, long long> un_mp1;
    for (long long i = 0; i < n; i++) {
      un_mp1[i] = -1;
    }
    right_next_greater_stk.push(0);
    for (long long i = 1; i < n; i++) {
      if (!right_next_greater_stk.empty() &&
          v[right_next_greater_stk.top()] < v[i]) {
        while (!right_next_greater_stk.empty() &&
               v[right_next_greater_stk.top()] < v[i]) {
          un_mp1[right_next_greater_stk.top()] = i;
          right_next_greater_stk.pop();
        }
        right_next_greater_stk.push(i);
      } else {
        right_next_greater_stk.push(i);
      }
    }
    unordered_map<long long, long long> un_mp2;
    for (long long i = 0; i < n; i++) {
      un_mp2[i] = -1;
    }
    left_next_greater_stk.push(n - 1);
    for (long long i = n - 1; i >= 0; i--) {
      if (!left_next_greater_stk.empty() &&
          v[left_next_greater_stk.top()] < v[i]) {
        while (!left_next_greater_stk.empty() &&
               v[left_next_greater_stk.top()] < v[i]) {
          un_mp2[left_next_greater_stk.top()] = i;
          left_next_greater_stk.pop();
        }
        left_next_greater_stk.push(i);
      } else {
        left_next_greater_stk.push(i);
      }
    }
    long long maxi = -1000000007;
    for (auto x : un_mp1) {
      if (x.second == -1) continue;
      maxi = max(maxi, (v[x.first] ^ v[x.second]));
    }
    for (auto x : un_mp2) {
      if (x.second == -1) continue;
      maxi = max(maxi, (v[x.first] ^ v[x.second]));
    }
    cout << maxi << "\n";
  }
  return 0;
}
