#include <bits/stdc++.h>
using namespace std;
int main(int argc, char *argv[]) {
  vector<int> numbers;
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int a;
    cin >> a;
    numbers.push_back(a);
  }
  set<int> highest;
  for (int h = 30; h >= 0; --h) {
    bool found = false;
    bool notfound = false;
    highest.clear();
    for (int i = 0; i < n; ++i) {
      if ((numbers[i] & (1 << h)) > 0) {
        found = true;
        highest.insert(i);
      } else {
        notfound = true;
      }
    }
    if (found && notfound) {
      int lucky = -1;
      set<int>::iterator it;
      for (it = highest.begin(); it != highest.end(); ++it) {
        int curr = *it;
        int currn = numbers[curr];
        int max = 0;
        for (int k = curr - 1; k >= 0; --k) {
          if (highest.find(k) != highest.end()) {
            break;
          }
          int sec = numbers[k];
          if (sec < max) {
            continue;
          } else {
            max = sec;
          }
          if ((currn ^ sec) > lucky) {
            lucky = currn ^ sec;
          }
        }
        max = 0;
        for (int k = curr + 1; k < n; ++k) {
          if (highest.find(k) != highest.end()) {
            break;
          }
          int sec = numbers[k];
          if (sec < max) {
            continue;
          } else {
            max = sec;
          }
          if ((currn ^ sec) > lucky) {
            lucky = currn ^ sec;
          }
        }
      }
      cout << lucky;
      return 0;
    }
  }
  return EXIT_SUCCESS;
}

### 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 main(int argc, char *argv[]) {
  vector<int> numbers;
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int a;
    cin >> a;
    numbers.push_back(a);
  }
  set<int> highest;
  for (int h = 30; h >= 0; --h) {
    bool found = false;
    bool notfound = false;
    highest.clear();
    for (int i = 0; i < n; ++i) {
      if ((numbers[i] & (1 << h)) > 0) {
        found = true;
        highest.insert(i);
      } else {
        notfound = true;
      }
    }
    if (found && notfound) {
      int lucky = -1;
      set<int>::iterator it;
      for (it = highest.begin(); it != highest.end(); ++it) {
        int curr = *it;
        int currn = numbers[curr];
        int max = 0;
        for (int k = curr - 1; k >= 0; --k) {
          if (highest.find(k) != highest.end()) {
            break;
          }
          int sec = numbers[k];
          if (sec < max) {
            continue;
          } else {
            max = sec;
          }
          if ((currn ^ sec) > lucky) {
            lucky = currn ^ sec;
          }
        }
        max = 0;
        for (int k = curr + 1; k < n; ++k) {
          if (highest.find(k) != highest.end()) {
            break;
          }
          int sec = numbers[k];
          if (sec < max) {
            continue;
          } else {
            max = sec;
          }
          if ((currn ^ sec) > lucky) {
            lucky = currn ^ sec;
          }
        }
      }
      cout << lucky;
      return 0;
    }
  }
  return EXIT_SUCCESS;
}

```