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}.
#include <bits/stdc++.h>
using namespace std;
int main() {
  int i, j, n;
  long long int ans = 0;
  scanf("%d", &n);
  int arr[n], next[n], prev[n];
  for (i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
  }
  prev[0] = -1;
  stack<int> s;
  s.push(0);
  for (i = 1; i < n; i++) {
    while (s.empty() != true && arr[s.top()] < arr[i]) {
      s.pop();
    }
    if (s.empty() == true) {
      prev[i] = -1;
    } else {
      prev[i] = s.top();
    }
    s.push(i);
  }
  next[n - 1] = n;
  stack<int> q;
  q.push(n - 1);
  for (i = n - 2; i >= 0; i--) {
    while (q.empty() != true && arr[q.top()] < arr[i]) {
      q.pop();
    }
    if (q.empty() == true) {
      next[i] = n;
    } else {
      next[i] = q.top();
    }
    q.push(i);
  }
  ans = INT_MIN;
  for (i = 0; i < n; i++) {
    if (prev[i] >= 0) {
      ans = max(ans, (long long int)(arr[i] ^ arr[(prev[i])]));
    }
    if (next[i] < n) {
      ans = max(ans, (long long int)(arr[i] ^ arr[(next[i])]));
    }
  }
  cout << ans << "\n";
}
