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 solve(long long int n) {
  long long int a[n];
  for (long long int i = 0; i < n; i++) cin >> a[i];
  stack<long long int> s;
  long long int xr = a[0] ^ a[1];
  s.push(0);
  s.push(1);
  for (long long int i = 2; i < n; i++) {
    if (a[s.top()] >= a[i]) {
      long long int f = a[s.top()] ^ a[i];
      xr = xr > f ? xr : f;
      s.push(i);
    } else {
      while (!s.empty() && a[s.top()] < a[i]) {
        long long int f = a[s.top()] ^ a[i];
        s.pop();
        xr = xr > f ? xr : f;
      }
      if (!s.empty()) {
        long long int f = a[s.top()] ^ a[i];
        xr = xr > f ? xr : f;
      }
      s.push(i);
    }
  }
  long long int ai = a[s.top()];
  s.pop();
  while (!s.empty()) {
    long long int f = ai ^ a[s.top()];
    xr = xr > f ? xr : f;
    s.pop();
    if (!s.empty() && ai < a[s.top()]) break;
  }
  cout << xr << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  cin >> t;
  solve(t);
  return 0;
}
