#include <bits/stdc++.h>
using namespace std;
int solve() {
  int n;
  cin >> n;
  vector<int> s(n);
  for (int i = 0; i < (n); i++) cin >> s[i];
  vector<int> next(n, -1);
  vector<int> prev(n, -1);
  stack<int> undetecteds;
  for (int i = 0; i < (n); i++) {
    while (!undetecteds.empty()) {
      int sti = undetecteds.top();
      if (s[sti] < s[i]) {
        next[sti] = i;
        undetecteds.pop();
      } else
        break;
    }
    undetecteds.push(i);
  }
  undetecteds = stack<int>();
  for (int i = n - 1; i >= 0; i--) {
    while (!undetecteds.empty()) {
      int sti = undetecteds.top();
      if (s[sti] < s[i]) {
        prev[sti] = i;
        undetecteds.pop();
      } else
        break;
    }
    undetecteds.push(i);
  }
  int maxXor = -1;
  for (int i = 0; i < (n); i++) {
    if (next[i] != -1) maxXor = max(maxXor, s[i] ^ s[next[i]]);
    if (prev[i] != -1) maxXor = max(maxXor, s[i] ^ s[prev[i]]);
  }
  return maxXor;
}
int main() {
  ios::sync_with_stdio(false);
  cout << solve() << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int solve() {
  int n;
  cin >> n;
  vector<int> s(n);
  for (int i = 0; i < (n); i++) cin >> s[i];
  vector<int> next(n, -1);
  vector<int> prev(n, -1);
  stack<int> undetecteds;
  for (int i = 0; i < (n); i++) {
    while (!undetecteds.empty()) {
      int sti = undetecteds.top();
      if (s[sti] < s[i]) {
        next[sti] = i;
        undetecteds.pop();
      } else
        break;
    }
    undetecteds.push(i);
  }
  undetecteds = stack<int>();
  for (int i = n - 1; i >= 0; i--) {
    while (!undetecteds.empty()) {
      int sti = undetecteds.top();
      if (s[sti] < s[i]) {
        prev[sti] = i;
        undetecteds.pop();
      } else
        break;
    }
    undetecteds.push(i);
  }
  int maxXor = -1;
  for (int i = 0; i < (n); i++) {
    if (next[i] != -1) maxXor = max(maxXor, s[i] ^ s[next[i]]);
    if (prev[i] != -1) maxXor = max(maxXor, s[i] ^ s[prev[i]]);
  }
  return maxXor;
}
int main() {
  ios::sync_with_stdio(false);
  cout << solve() << endl;
  return 0;
}

```