#include <bits/stdc++.h>
using namespace std;
long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long min(long long a, long long b) {
  if (a > b) return b;
  return a;
}
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
long long solve(vector<long long> arr) {
  long long ans = arr[0] ^ arr[1];
  vector<long long> ss;
  for (int i = 0; i < ((int)(arr).size()); i++) {
    while (!ss.empty() && ss.back() < arr[i]) {
      if (((int)(ss).size()) >= 2)
        ans = max(ans, ss[((int)(ss).size()) - 1] ^ ss[((int)(ss).size()) - 2]);
      ss.pop_back();
    }
    ss.push_back(arr[i]);
    if (((int)(ss).size()) >= 2)
      ans = max(ans, ss[((int)(ss).size()) - 1] ^ ss[((int)(ss).size()) - 2]);
  }
  return ans;
}
int main() {
  long long n;
  cin >> n;
  vector<long long> arr;
  for (int i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    arr.push_back(x);
  }
  long long ans = solve(arr);
  reverse(arr.begin(), arr.end());
  ans = max(ans, solve(arr));
  cout << ans << endl;
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

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}.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long min(long long a, long long b) {
  if (a > b) return b;
  return a;
}
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
long long solve(vector<long long> arr) {
  long long ans = arr[0] ^ arr[1];
  vector<long long> ss;
  for (int i = 0; i < ((int)(arr).size()); i++) {
    while (!ss.empty() && ss.back() < arr[i]) {
      if (((int)(ss).size()) >= 2)
        ans = max(ans, ss[((int)(ss).size()) - 1] ^ ss[((int)(ss).size()) - 2]);
      ss.pop_back();
    }
    ss.push_back(arr[i]);
    if (((int)(ss).size()) >= 2)
      ans = max(ans, ss[((int)(ss).size()) - 1] ^ ss[((int)(ss).size()) - 2]);
  }
  return ans;
}
int main() {
  long long n;
  cin >> n;
  vector<long long> arr;
  for (int i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    arr.push_back(x);
  }
  long long ans = solve(arr);
  reverse(arr.begin(), arr.end());
  ans = max(ans, solve(arr));
  cout << ans << endl;
  return 0;
}

```