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;
const int N = 100010;
vector<long long> is_prime(N + 1, true);
long long to_int(string s) {
  stringstream geek(s);
  long long x = 0;
  geek >> x;
  return x;
}
string to_string(long long n) {
  ostringstream str1;
  str1 << n;
  string geek = str1.str();
  return geek;
}
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
bool sortbysecdesc(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second > b.second;
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
long long powerm(long long a, long long k) {
  if (!k) {
    return 1;
  }
  long long b = powerm(a, k / 2);
  b = b * b % (1000000000 + 7);
  if (k % 2) {
    return a * b % (1000000000 + 7);
  } else {
    return b;
  }
}
long long power(long long a, long long b) {
  if (b == 1) return a;
  if (b == 0) return 1;
  long long m1 = power(a, b / 2);
  if (b % 2) return m1 * m1 * a;
  return m1 * m1;
}
void seive_of_eras(long long n) {
  is_prime[0] = 0;
  is_prime[1] = 0;
  for (long long i = 2; i * i <= n; i++) {
    if (is_prime[i]) {
      for (long long j = i * i; j <= n; j += i) {
        is_prime[j] = false;
      }
    }
  }
}
bool isprime(long long a) {
  if (a <= 1) return false;
  if (a == 2 | a == 3) return true;
  if (a % 2 == 0 || a % 3 == 0) return false;
  for (long long i = 5; i * i <= a; i = i + 6) {
    if (a % i == 0 || a % (i + 2) == 0) return false;
  }
  return true;
}
void codeit() {
  long long n;
  cin >> n;
  long long a[n];
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  stack<long long> s;
  for (long long i = 0; i < n; i++) {
    long long x = a[i];
    long long y;
    while (!s.empty()) {
      ans = max(ans, s.top() ^ x);
      if (x < s.top()) {
        break;
      } else {
        s.pop();
      }
    }
    s.push(x);
  }
  cout << ans << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  long long tt;
  tt = 1;
  for (long long i = 0; i < tt; i++) {
    codeit();
  }
}
