#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  int answer = 0;
  vector<int> stk;
  for (; n > 0; n--) {
    int x;
    cin >> x;
    while (not stk.empty() and stk.back() <= x) {
      answer = max(answer, stk.back() ^ x);
      stk.pop_back();
    }
    stk.push_back(x);
    if (stk.size() >= 2) {
      answer = max(answer, stk[stk.size() - 2] ^ stk[stk.size() - 1]);
    }
  }
  cout << answer << endl;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
int main() {
  int n;
  cin >> n;
  int answer = 0;
  vector<int> stk;
  for (; n > 0; n--) {
    int x;
    cin >> x;
    while (not stk.empty() and stk.back() <= x) {
      answer = max(answer, stk.back() ^ x);
      stk.pop_back();
    }
    stk.push_back(x);
    if (stk.size() >= 2) {
      answer = max(answer, stk[stk.size() - 2] ^ stk[stk.size() - 1]);
    }
  }
  cout << answer << endl;
}

```