#include <bits/stdc++.h>
using namespace std;
const int nax = 1000 * 1007;
const int INF = 1e9;
long long int pow(long long int x, long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  vector<int> s;
  int ans = v[0] ^ v[1];
  for (int i = 0; i < n; i++) {
    while (s.size() != 0 && s.back() < v[i]) s.pop_back();
    s.push_back(v[i]);
    if (s.size() >= 2) ans = max(ans, s[s.size() - 1] ^ s[s.size() - 2]);
  }
  s.clear();
  reverse(v.begin(), v.end());
  ans = max(ans, v[0] ^ v[1]);
  for (int i = 0; i < n; i++) {
    while (s.size() != 0 && s.back() < v[i]) s.pop_back();
    s.push_back(v[i]);
    if (s.size() >= 2) ans = max(ans, s[s.size() - 1] ^ s[s.size() - 2]);
  }
  cout << ans << endl;
}

### Prompt

Develop a solution in CPP 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 nax = 1000 * 1007;
const int INF = 1e9;
long long int pow(long long int x, long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  vector<int> s;
  int ans = v[0] ^ v[1];
  for (int i = 0; i < n; i++) {
    while (s.size() != 0 && s.back() < v[i]) s.pop_back();
    s.push_back(v[i]);
    if (s.size() >= 2) ans = max(ans, s[s.size() - 1] ^ s[s.size() - 2]);
  }
  s.clear();
  reverse(v.begin(), v.end());
  ans = max(ans, v[0] ^ v[1]);
  for (int i = 0; i < n; i++) {
    while (s.size() != 0 && s.back() < v[i]) s.pop_back();
    s.push_back(v[i]);
    if (s.size() >= 2) ans = max(ans, s[s.size() - 1] ^ s[s.size() - 2]);
  }
  cout << ans << endl;
}

```