#include <bits/stdc++.h>
using namespace std;
long long arr[100009];
vector<long long> lock;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  if (n == 2) {
    cout << (arr[1] ^ arr[2]) << endl;
    return 0;
  }
  long long ans = 0;
  stack<long long> msla;
  msla.push(arr[1]);
  for (int i = 2; i <= n; i++) {
    long long cuel = arr[i];
    ans = max(ans, (arr[i] ^ arr[i - 1]));
    long long kala = -1;
    while (true) {
      if (msla.empty() || (msla.top() > cuel)) {
        if (!msla.empty()) kala = msla.top();
        msla.push(cuel);
        break;
      }
      long long yopa = msla.top();
      if (yopa < cuel) {
        msla.pop();
      }
    }
    if (kala != -1) {
      ans = max(ans, cuel ^ kala);
    }
  }
  stack<long long> msla2;
  msla2.push(arr[n]);
  for (int i = n - 1; i >= 1; i--) {
    long long cuel = arr[i];
    ans = max(ans, (arr[i] ^ arr[i + 1]));
    long long kala = -1;
    while (true) {
      if (msla2.empty() || (msla2.top() > cuel)) {
        if (!msla2.empty()) kala = msla2.top();
        msla2.push(cuel);
        break;
      }
      long long yopa = msla2.top();
      if (yopa < cuel) {
        msla2.pop();
      }
    }
    if (kala != -1) {
      ans = max(ans, cuel ^ kala);
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
long long arr[100009];
vector<long long> lock;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  if (n == 2) {
    cout << (arr[1] ^ arr[2]) << endl;
    return 0;
  }
  long long ans = 0;
  stack<long long> msla;
  msla.push(arr[1]);
  for (int i = 2; i <= n; i++) {
    long long cuel = arr[i];
    ans = max(ans, (arr[i] ^ arr[i - 1]));
    long long kala = -1;
    while (true) {
      if (msla.empty() || (msla.top() > cuel)) {
        if (!msla.empty()) kala = msla.top();
        msla.push(cuel);
        break;
      }
      long long yopa = msla.top();
      if (yopa < cuel) {
        msla.pop();
      }
    }
    if (kala != -1) {
      ans = max(ans, cuel ^ kala);
    }
  }
  stack<long long> msla2;
  msla2.push(arr[n]);
  for (int i = n - 1; i >= 1; i--) {
    long long cuel = arr[i];
    ans = max(ans, (arr[i] ^ arr[i + 1]));
    long long kala = -1;
    while (true) {
      if (msla2.empty() || (msla2.top() > cuel)) {
        if (!msla2.empty()) kala = msla2.top();
        msla2.push(cuel);
        break;
      }
      long long yopa = msla2.top();
      if (yopa < cuel) {
        msla2.pop();
      }
    }
    if (kala != -1) {
      ans = max(ans, cuel ^ kala);
    }
  }
  cout << ans << endl;
  return 0;
}

```