#include <bits/stdc++.h>
using namespace std;
int main() {
  int len;
  cin >> len;
  int inp[len], stack[len], nxt[len], prev[len], head = -1;
  for (int i = 0; i < len; ++i) {
    cin >> inp[i];
    if (head != -1 && inp[stack[head]] < inp[i]) {
      while (head >= 0 && inp[stack[head]] < inp[i]) nxt[stack[head--]] = i;
    }
    stack[++head] = i;
  }
  while (head > -1) nxt[stack[head--]] = -1;
  for (int i = len - 1; i > -1; --i) {
    if (head != -1 && inp[stack[head]] < inp[i]) {
      while (head >= 0 && inp[stack[head]] < inp[i]) prev[stack[head--]] = i;
    }
    stack[++head] = i;
  }
  while (head > -1) prev[stack[head--]] = -1;
  int max = 0;
  for (int i = 0; i < len; ++i) {
    int temp;
    if (nxt[i] != -1) {
      temp = inp[i] ^ inp[nxt[i]];
      max = ((temp > max) ? temp : max);
    }
    if (prev[i] != -1) {
      temp = inp[i] ^ inp[prev[i]];
      max = ((temp > max) ? temp : max);
    }
  }
  cout << max << '\n';
}

### Prompt

Your challenge is to write a CPP solution to 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 main() {
  int len;
  cin >> len;
  int inp[len], stack[len], nxt[len], prev[len], head = -1;
  for (int i = 0; i < len; ++i) {
    cin >> inp[i];
    if (head != -1 && inp[stack[head]] < inp[i]) {
      while (head >= 0 && inp[stack[head]] < inp[i]) nxt[stack[head--]] = i;
    }
    stack[++head] = i;
  }
  while (head > -1) nxt[stack[head--]] = -1;
  for (int i = len - 1; i > -1; --i) {
    if (head != -1 && inp[stack[head]] < inp[i]) {
      while (head >= 0 && inp[stack[head]] < inp[i]) prev[stack[head--]] = i;
    }
    stack[++head] = i;
  }
  while (head > -1) prev[stack[head--]] = -1;
  int max = 0;
  for (int i = 0; i < len; ++i) {
    int temp;
    if (nxt[i] != -1) {
      temp = inp[i] ^ inp[nxt[i]];
      max = ((temp > max) ? temp : max);
    }
    if (prev[i] != -1) {
      temp = inp[i] ^ inp[prev[i]];
      max = ((temp > max) ? temp : max);
    }
  }
  cout << max << '\n';
}

```