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;
using ll = long long;
ll dx[] = {0, 0, 1, -1};
ll dy[] = {1, -1, 0, 0};
void solve() {
  ll n;
  cin >> n;
  vector<ll> v(n);
  for (ll i = 0; i < (n); ++i) cin >> v[i];
  vector<pair<ll, ll>> aux(n);
  ll i = n - 1;
  stack<ll> s;
  while (i >= 0) {
    while (!s.empty() && s.top() < v[i]) s.pop();
    if (s.empty())
      aux[i].first = -1;
    else
      aux[i].first = s.top();
    s.push(v[i]);
    i--;
  }
  i = 0;
  while (!s.empty()) s.pop();
  while (i < n) {
    while (!s.empty() && s.top() < v[i]) s.pop();
    if (s.empty())
      aux[i].second = -1;
    else
      aux[i].second = s.top();
    s.push(v[i]), i++;
  }
  ll res = 0;
  for (ll i = 0; i < (n); ++i) {
    if (aux[i].first != -1) res = max(res, aux[i].first ^ v[i]);
    if (aux[i].second != -1) res = max(res, aux[i].second ^ v[i]);
  }
  cout << res << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  ll t = 1;
  while (t--) solve();
  return 0;
}
