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() {
  long long n;
  cin >> n;
  vector<long long> A(n);
  vector<long long> L(n, -1);
  vector<long long> R(n, -1);
  stack<long long> s;
  for (long long i = 0; i < n; i++) cin >> A[i];
  for (long long i = 0; i < n; i++) {
    while (!s.empty() && A[s.top()] < A[i]) {
      long long ind = s.top();
      s.pop();
      L[ind] = A[i];
    }
    s.push(i);
  }
  while (!s.empty()) s.pop();
  for (long long i = n - 1; i >= 0; i--) {
    while (!s.empty() && A[s.top()] < A[i]) {
      long long ind = s.top();
      s.pop();
      R[ind] = A[i];
    }
    s.push(i);
  }
  long long mx = 0;
  for (long long i = 0; i < n; i++) {
    long long mx1 = L[i] == -1 ? 0 : A[i] ^ L[i];
    long long mx2 = R[i] == -1 ? 0 : A[i] ^ R[i];
    mx = max(mx, max(mx1, mx2));
  }
  cout << mx << endl;
  return 0;
}
