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 = 100005;
void solve() {
  long long int n;
  cin >> n;
  long long int a[n];
  long long int ans = 0;
  long long int left[n], right[n];
  memset(left, -1, sizeof(left));
  stack<long long int> s;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    right[i] = n;
    while (!s.empty() && a[s.top()] < a[i]) s.pop();
    if (!s.empty()) left[i] = s.top();
    s.push(i);
  }
  while (!s.empty()) s.pop();
  for (long long int i = n - 1; i >= 0; i--) {
    while (!s.empty() && a[s.top()] < a[i]) s.pop();
    if (!s.empty()) right[i] = s.top();
    s.push(i);
  }
  for (long long int i = 0; i < n; i++) {
    long long int start = left[i] + 1;
    long long int end = right[i] - 1;
    for (long long int j = i - 1; j >= start; j = left[j]) {
      ans = max(ans, a[i] ^ a[j]);
    }
    for (long long int j = i + 1; j <= end; j = right[j]) {
      ans = max(ans, a[i] ^ a[j]);
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
}
