#include <bits/stdc++.h>
using namespace std;
void runCase(int caseNum) {
  int n;
  cin >> n;
  vector<int> nums;
  vector<pair<int, int> > pos;
  vector<int> left(n, 0), right(n, 0), leftmark(n, 0), rightmark(n, 0);
  set<int> ss;
  for (int i = 0; i < n; ++i) {
    int t;
    cin >> t;
    nums.push_back(t);
    pos.push_back(make_pair(t, i));
  }
  sort(pos.begin(), pos.end());
  reverse(pos.begin(), pos.end());
  for (vector<pair<int, int> >::iterator it = pos.begin(); it != pos.end();
       ++it) {
    int a = (*it).second;
    if (ss.empty()) {
      ss.insert(a);
      left[a] = right[a] = -1;
      leftmark[a] = rightmark[a] = 1;
      continue;
    }
    set<int>::iterator p = ss.lower_bound(a);
    if (p == ss.end()) {
      right[a] = -1;
      rightmark[a] = 1;
    } else
      right[a] = *p;
    if (p == ss.begin()) {
      left[a] = -1;
      leftmark[a] = 1;
    } else
      left[a] = *(--p);
    ss.insert(a);
  }
  reverse(pos.begin(), pos.end());
  int res = 0;
  for (int i = 0; i < n; ++i) {
    int cur = pos[i].second;
    while (cur != -1) {
      if (leftmark[cur]) break;
      res = max(res, nums[left[cur]] ^ nums[cur]);
      leftmark[cur] = 1;
      cur = left[cur];
    }
    cur = pos[i].second;
    while (cur != -1) {
      if (rightmark[cur]) break;
      res = max(res, nums[right[cur]] ^ nums[cur]);
      rightmark[cur] = 1;
      cur = right[cur];
    }
  }
  cout << res << endl;
}
int main(int argc, char* argv[]) {
  runCase(0);
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
void runCase(int caseNum) {
  int n;
  cin >> n;
  vector<int> nums;
  vector<pair<int, int> > pos;
  vector<int> left(n, 0), right(n, 0), leftmark(n, 0), rightmark(n, 0);
  set<int> ss;
  for (int i = 0; i < n; ++i) {
    int t;
    cin >> t;
    nums.push_back(t);
    pos.push_back(make_pair(t, i));
  }
  sort(pos.begin(), pos.end());
  reverse(pos.begin(), pos.end());
  for (vector<pair<int, int> >::iterator it = pos.begin(); it != pos.end();
       ++it) {
    int a = (*it).second;
    if (ss.empty()) {
      ss.insert(a);
      left[a] = right[a] = -1;
      leftmark[a] = rightmark[a] = 1;
      continue;
    }
    set<int>::iterator p = ss.lower_bound(a);
    if (p == ss.end()) {
      right[a] = -1;
      rightmark[a] = 1;
    } else
      right[a] = *p;
    if (p == ss.begin()) {
      left[a] = -1;
      leftmark[a] = 1;
    } else
      left[a] = *(--p);
    ss.insert(a);
  }
  reverse(pos.begin(), pos.end());
  int res = 0;
  for (int i = 0; i < n; ++i) {
    int cur = pos[i].second;
    while (cur != -1) {
      if (leftmark[cur]) break;
      res = max(res, nums[left[cur]] ^ nums[cur]);
      leftmark[cur] = 1;
      cur = left[cur];
    }
    cur = pos[i].second;
    while (cur != -1) {
      if (rightmark[cur]) break;
      res = max(res, nums[right[cur]] ^ nums[cur]);
      rightmark[cur] = 1;
      cur = right[cur];
    }
  }
  cout << res << endl;
}
int main(int argc, char* argv[]) {
  runCase(0);
  return 0;
}

```