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 arr[100005], nxt[100005];
long long n;
void nxt_greater() {
  stack<long long> s;
  s.push(0);
  for (long long i = 1; i < n; i++) {
    while (!s.empty() && arr[s.top()] < arr[i]) {
      nxt[s.top()] = i;
      s.pop();
    }
    s.push(i);
  }
  while (!s.empty()) {
    nxt[s.top()] = -1;
    s.pop();
  }
}
int main() {
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> arr[i];
  nxt_greater();
  long long max_xor = 0;
  for (long long i = 0; i < n; i++) {
    if (nxt[i] >= 0) {
      max_xor = max(max_xor, arr[i] ^ arr[nxt[i]]);
    }
  }
  reverse(arr, arr + n);
  nxt_greater();
  for (long long i = 0; i < n; i++) {
    if (nxt[i] >= 0) {
      max_xor = max(max_xor, arr[i] ^ arr[nxt[i]]);
    }
  }
  cout << max_xor << endl;
}
