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;
class Stack {
  int top;

 public:
  long long a[100000];
  Stack() { top = -1; }
  void push(long long x);
  long long pop();
  long long peek();
  bool isEmpty() { return (top == -1); }
  bool isFull() { return (top == (100000 - 1)); }
};
void Stack::push(long long x) {
  if (isFull()) return;
  a[++top] = x;
}
long long Stack::pop() {
  if (isEmpty()) return -1;
  return a[top--];
}
long long Stack::peek() { return a[top]; }
long long Xor(long long x, long long y) { return (x ^ y); }
int main() {
  int n;
  cin >> n;
  long long* arr = new long long[n];
  for (int i = 0; i < n; ++i) cin >> arr[i];
  Stack stck;
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    while (!stck.isEmpty()) {
      if (stck.peek() < arr[i]) {
        ans = max(ans, Xor(stck.pop(), arr[i]));
      } else {
        ans = max(ans, Xor(stck.peek(), arr[i]));
        break;
      }
    }
    stck.push(arr[i]);
  }
  cout << ans << endl;
}
