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>
T gcd(T a, T b) {
  return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
long long pwr(long long base, long long p) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % (1000000007LL);
    base = (base * base) % (1000000007LL);
    p >>= 1;
  }
  return ans;
}
long long modInv(long long n, long long mod) { return pwr(n, mod - 2); }
int bitCount(int n) {
  int c = 0;
  while (n) {
    c++;
    n &= n - 1;
  }
  return c;
}
int arr[200000];
stack<long long> st;
int main() {
  std::ios::sync_with_stdio(false);
  int n, i, temp;
  cin >> n;
  int p = 0, ans = 0;
  for (i = 0; i < n; i++) {
    cin >> temp;
    while (p > 0) {
      ans = max(ans, temp ^ arr[p]);
      ;
      if (temp > arr[p])
        p--;
      else
        break;
    }
    arr[++p] = temp;
  }
  cout << ans;
  return 0;
}
