#include <bits/stdc++.h>
using namespace std;
vector<pair<long long, long long> > m;
void NGE(long long a[], long long n) {
  stack<int> s;
  s.push(a[0]);
  for (long long i = 1; i < n; i++) {
    int next = a[i];
    if (s.empty() == false) {
      long long element = s.top();
      s.pop();
      while (element < next) {
        m.push_back({element, next});
        if (s.empty() == true) break;
        element = s.top();
        s.pop();
      }
      if (element > next) s.push(element);
    }
    s.push(next);
  }
  while (s.empty() == false) {
    m.push_back({s.top(), -1});
    s.pop();
  }
}
int main() {
  long long a[100002], n;
  cin >> n;
  long long xo, maxi = 0;
  for (long long i = 0; i < n; i++) cin >> a[i];
  NGE(a, n);
  reverse(a, a + n);
  NGE(a, n);
  for (long long i = 0; i < m.size(); i++)
    if (m[i].second != '1') {
      xo = m[i].first ^ m[i].second;
      if (xo > maxi) maxi = xo;
    }
  cout << maxi;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
vector<pair<long long, long long> > m;
void NGE(long long a[], long long n) {
  stack<int> s;
  s.push(a[0]);
  for (long long i = 1; i < n; i++) {
    int next = a[i];
    if (s.empty() == false) {
      long long element = s.top();
      s.pop();
      while (element < next) {
        m.push_back({element, next});
        if (s.empty() == true) break;
        element = s.top();
        s.pop();
      }
      if (element > next) s.push(element);
    }
    s.push(next);
  }
  while (s.empty() == false) {
    m.push_back({s.top(), -1});
    s.pop();
  }
}
int main() {
  long long a[100002], n;
  cin >> n;
  long long xo, maxi = 0;
  for (long long i = 0; i < n; i++) cin >> a[i];
  NGE(a, n);
  reverse(a, a + n);
  NGE(a, n);
  for (long long i = 0; i < m.size(); i++)
    if (m[i].second != '1') {
      xo = m[i].first ^ m[i].second;
      if (xo > maxi) maxi = xo;
    }
  cout << maxi;
}

```