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;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool isperfect(long long n) {
  long long y = sqrt(n);
  if (n % y == 0 && y * y == n) return true;
  return false;
}
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.second == b.second) return a.first > b.first;
  return a.second < b.second;
}
long long powi(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = ans * a;
    b = b >> 1;
    a = a * a;
  }
  return ans;
}
bool isprime(long long n) {
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<long long int> v(n);
  for (long long i = 0; i < n; i++) cin >> v[i];
  stack<long long> s;
  s.push(v[0]);
  long long ans = -1;
  for (long long i = 1; i < n; i++) {
    while (!s.empty() && s.top() < v[i]) {
      long long x = s.top();
      ans = max(ans, v[i] ^ x);
      s.pop();
    }
    if (!s.empty()) {
      int x = s.top();
      ans = max(ans, x ^ v[i]);
    }
    s.push(v[i]);
  }
  cout << ans;
}
