#include <bits/stdc++.h>
using namespace std;
int n;
int a[100005];
int less_from_left[100005], less_from_right[100005];
int greater_from_left[100005], greater_from_right[100005];
map<int, int> M;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < (n); ++i) cin >> a[i], M[a[i]] = i;
  deque<int> deq;
  less_from_left[0] = a[0];
  deq.push_front(a[0]);
  for (int i = 1; i <= (n - 1); ++i) {
    while (!deq.empty() && deq.front() > a[i]) deq.pop_front();
    less_from_left[i] = deq.empty() ? a[0] : deq.front();
    deq.push_front(a[i]);
  }
  deq.clear();
  greater_from_left[0] = a[0];
  deq.push_front(a[0]);
  for (int i = 1; i <= (n - 1); ++i) {
    while (!deq.empty() && deq.front() < a[i]) deq.pop_front();
    greater_from_left[i] = deq.empty() ? a[i] : deq.front();
    deq.push_front(a[i]);
  }
  deq.clear();
  less_from_right[n - 1] = a[n - 1];
  deq.push_front(a[n - 1]);
  for (int i = n - 2; i >= (0); --i) {
    while (!deq.empty() && deq.front() > a[i]) deq.pop_front();
    less_from_right[i] = deq.empty() ? a[n - 1] : deq.front();
    deq.push_front(a[i]);
  }
  deq.clear();
  greater_from_right[n - 1] = a[n - 1];
  deq.push_front(a[n - 1]);
  for (int i = n - 2; i >= (0); --i) {
    while (!deq.empty() && deq.front() < a[i]) deq.pop_front();
    greater_from_right[i] = deq.empty() ? a[i] : deq.front();
    deq.push_front(a[i]);
  }
  int w = 0;
  for (int i = 0; i < (n); ++i) {
    w = max(w, a[i] ^ greater_from_left[i]);
    w = max(w, a[i] ^ greater_from_right[i]);
  }
  cout << w << "\n";
  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 n;
int a[100005];
int less_from_left[100005], less_from_right[100005];
int greater_from_left[100005], greater_from_right[100005];
map<int, int> M;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < (n); ++i) cin >> a[i], M[a[i]] = i;
  deque<int> deq;
  less_from_left[0] = a[0];
  deq.push_front(a[0]);
  for (int i = 1; i <= (n - 1); ++i) {
    while (!deq.empty() && deq.front() > a[i]) deq.pop_front();
    less_from_left[i] = deq.empty() ? a[0] : deq.front();
    deq.push_front(a[i]);
  }
  deq.clear();
  greater_from_left[0] = a[0];
  deq.push_front(a[0]);
  for (int i = 1; i <= (n - 1); ++i) {
    while (!deq.empty() && deq.front() < a[i]) deq.pop_front();
    greater_from_left[i] = deq.empty() ? a[i] : deq.front();
    deq.push_front(a[i]);
  }
  deq.clear();
  less_from_right[n - 1] = a[n - 1];
  deq.push_front(a[n - 1]);
  for (int i = n - 2; i >= (0); --i) {
    while (!deq.empty() && deq.front() > a[i]) deq.pop_front();
    less_from_right[i] = deq.empty() ? a[n - 1] : deq.front();
    deq.push_front(a[i]);
  }
  deq.clear();
  greater_from_right[n - 1] = a[n - 1];
  deq.push_front(a[n - 1]);
  for (int i = n - 2; i >= (0); --i) {
    while (!deq.empty() && deq.front() < a[i]) deq.pop_front();
    greater_from_right[i] = deq.empty() ? a[i] : deq.front();
    deq.push_front(a[i]);
  }
  int w = 0;
  for (int i = 0; i < (n); ++i) {
    w = max(w, a[i] ^ greater_from_left[i]);
    w = max(w, a[i] ^ greater_from_right[i]);
  }
  cout << w << "\n";
  return 0;
}

```