#include <bits/stdc++.h>
using namespace std;
const int INF = (1LL << 31) - 1;
const int maxn = 1e5;
vector<int> t;
int n, a[maxn], ans;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    while (t.size() && t.back() < a[i])
      ans = max(ans, a[i] ^ t.back()), t.pop_back();
    if (t.size()) ans = max(ans, a[i] ^ t.back());
    t.push_back(a[i]);
  }
  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 INF = (1LL << 31) - 1;
const int maxn = 1e5;
vector<int> t;
int n, a[maxn], ans;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    while (t.size() && t.back() < a[i])
      ans = max(ans, a[i] ^ t.back()), t.pop_back();
    if (t.size()) ans = max(ans, a[i] ^ t.back());
    t.push_back(a[i]);
  }
  cout << ans << endl;
}

```