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;
void bit(unsigned long long int a, vector<int> &arr) {
  while (a > 0) {
    arr.push_back(a % 2);
    a /= 2;
  }
  reverse(arr.begin(), arr.end());
}
unsigned long long int power(unsigned long long int a, int b) {
  unsigned long long int res = 1;
  while (b > 0) {
    if (b & 1) res *= a;
    b = b >> 1;
    a *= a;
  }
  return res;
}
int n;
int main() {
  int n;
  cin >> n;
  vector<long long> a(n, 0 * 1LL);
  for (int i = 0; i < n; i++) cin >> a[i];
  stack<long long> second;
  long long ans = -1;
  second.push(a[0]);
  int i = 1;
  while (i < n) {
    if (a[i] < second.top()) {
      ans = max(ans, a[i] ^ second.top());
    } else {
      while (!second.empty() && a[i] > second.top()) {
        ans = max(ans, a[i] ^ second.top());
        second.pop();
      }
      if (!second.empty()) ans = max(ans, a[i] ^ second.top());
    }
    second.push(a[i]);
    i++;
  }
  cout << ans << "\n";
}
