#include <bits/stdc++.h>
using namespace std;
const int dir[4][2] = {1, 0, 0, 1, -1, 0, 0, -1};
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
const double pi = acos(-1);
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
int a[N];
int n, m, K;
stack<int> s;
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  vector<int> res;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    while (!s.empty() && a[i] > s.top()) s.pop();
    if (!s.empty()) res.push_back(a[i] ^ s.top());
    s.push(a[i]);
  }
  while (!s.empty()) s.pop();
  for (int i = n; i; --i) {
    while (!s.empty() && a[i] > s.top()) s.pop();
    if (!s.empty()) res.push_back(a[i] ^ s.top());
    s.push(a[i]);
  }
  sort(res.begin(), res.end());
  cout << res[res.size() - 1] << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const int dir[4][2] = {1, 0, 0, 1, -1, 0, 0, -1};
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
const double pi = acos(-1);
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
int a[N];
int n, m, K;
stack<int> s;
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  vector<int> res;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    while (!s.empty() && a[i] > s.top()) s.pop();
    if (!s.empty()) res.push_back(a[i] ^ s.top());
    s.push(a[i]);
  }
  while (!s.empty()) s.pop();
  for (int i = n; i; --i) {
    while (!s.empty() && a[i] > s.top()) s.pop();
    if (!s.empty()) res.push_back(a[i] ^ s.top());
    s.push(a[i]);
  }
  sort(res.begin(), res.end());
  cout << res[res.size() - 1] << endl;
  return 0;
}

```