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;
template <class T>
inline void ri(T &i) {
  bool minus = false;
  char c;
  for (c = getchar(); (c < '0' || c > '9') && (c != '-'); c = getchar())
    ;
  if (c == '-') {
    minus = true;
    c = '0';
  }
  for (i = 0; c >= '0' && c <= '9'; c = getchar())
    i = (i << 3) + (i << 1) + (c - 48);
  if (minus) i = (~i) + 1;
}
vector<long long> in;
int main() {
  int n;
  ri(n);
  long long t;
  for (int i = 0; i < n; i++) {
    ri(t);
    in.push_back(t);
  }
  long long ans = 0;
  vector<long long> tem;
  tem.push_back(in[0]);
  for (int i = 1; i < n; i++) {
    t = tem.size() - 1;
    long long cur = in[i];
    while (tem[t] < cur) {
      ans = max(ans, cur ^ tem[t]);
      tem.pop_back();
      t--;
    }
    if (tem.size()) {
      ans = max(ans, cur ^ tem[t]);
    }
    tem.push_back(cur);
  }
  cout << ans << endl;
  return 0;
}
