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;
long long n, arr[100001], satu[100001], maks, ans, oh;
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    long long tmp = arr[i];
    long long anu = 0;
    while (tmp > 0) {
      satu[anu] += tmp % 2;
      tmp /= 2;
      anu++;
    }
    maks = max(maks, anu - 1);
  }
  while (satu[maks] == n || satu[maks] == 0) maks--;
  int i = 0;
  int j = i + 1;
  oh = 0;
  while (i < n - 1) {
    if (j == n) break;
    if (!(arr[i] & (1 << maks)))
      i++, j = i + 1, oh = 0;
    else {
      oh = max(oh, arr[j]);
      if (arr[j] >= oh && arr[i] >= oh) ans = max(ans, arr[i] ^ arr[j]);
      if ((arr[j] & (1 << maks)))
        i++, j = i + 1, oh = 0;
      else
        j++;
    }
  }
  i = n;
  j = i - 1;
  oh = 0;
  while (i >= 1) {
    if (j == -1) break;
    if (!(arr[i] & (1 << maks)))
      i--, j = i - 1, oh = 0;
    else {
      oh = max(oh, arr[j]);
      if (arr[j] >= oh && arr[i] >= oh) ans = max(ans, arr[i] ^ arr[j]);
      if ((arr[j] & (1 << maks)))
        i--, j = i - 1, oh = 0;
      else
        j--;
    }
  }
  cout << ans << endl;
}
