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>
int diru[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dirv[] = {-1, 0, 1, -1, 1, -1, 0, 1};
using namespace std;
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
template <class T>
T sq(T n) {
  return n * n;
}
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);
}
template <class T>
bool inside(T a, T b, T c) {
  return a <= b && b <= c;
}
template <class T>
void setmax(T& a, T b) {
  if (a < b) a = b;
}
template <class T>
void setmin(T& a, T b) {
  if (b < a) a = b;
}
template <class T>
T power(T N, T P) {
  return (P == 0) ? 1 : N * power(N, P - 1);
}
int n;
int in[100010];
int q[100010];
int solve() {
  int len = 0, i, ret = 0;
  for (i = 0; i < n; i++) {
    while (len && in[i] > q[len - 1]) {
      ret = max(ret, in[i] ^ q[len - 1]);
      len--;
    }
    q[len++] = in[i];
  }
  return ret;
}
int main() {
  int T, t = 1, m, i, j, k;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &in[i]);
  m = solve();
  reverse(in, in + n);
  m = max(m, solve());
  printf("%d\n", m);
  return 0;
}
