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 solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<int> l(n), r(n);
  stack<pair<int, int>> s;
  for (int i = 0; i < n; i++) {
    while (s.size() && s.top().first < a[i]) {
      s.pop();
    }
    if (s.empty())
      l[i] = -1;
    else
      l[i] = s.top().second;
    s.push({a[i], i});
  }
  while (!s.empty()) {
    s.pop();
  }
  for (int i = n - 1; i >= 0; i--) {
    while (s.size() && s.top().first < a[i]) {
      s.pop();
    }
    if (s.empty())
      r[i] = n;
    else
      r[i] = s.top().second;
    s.push({a[i], i});
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j = r[j]) {
      if (a[j] >= a[i])
        break;
      else
        ans = max(ans, a[i] ^ a[j]);
    }
    for (int j = i - 1; j >= 0; j = l[j]) {
      if (a[j] >= a[i])
        break;
      else
        ans = max(ans, a[i] ^ a[j]);
    }
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q = 1;
  for (; q > 0; q--) {
    solve();
  }
}
