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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const double PI = 2.0 * acos(0.0);
const int MAXN = 100111;
int a[MAXN];
int n, res;
vector<int> s;
int main() {
  scanf("%d", &n);
  for (int i = 0, _n = (n); i < _n; i++) scanf("%d", &a[i]);
  vector<int> s;
  for (int i = 0, _n = (n); i < _n; i++) {
    while (!s.empty() && (a[i] > a[s.back()])) s.pop_back();
    if (!s.empty()) {
      int j = s.back();
      res = max(res, a[i] ^ a[j]);
    }
    s.push_back(i);
  }
  while (!s.empty()) s.pop_back();
  for (int i = (n)-1; i >= 0; i--) {
    while (!s.empty() && (a[i] > a[s.back()])) s.pop_back();
    if (!s.empty()) {
      int j = s.back();
      res = max(res, a[i] ^ a[j]);
    }
    s.push_back(i);
  }
  printf("%d\n", res);
}
