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;
struct stak {
  int t = -1;
  int arr[200500];
  int size() { return t + 1; }
  void push(int val) {
    t++;
    if (t < 200500) {
      arr[t] = val;
    } else
      printf("full\n");
  }
  void pop() {
    if (t >= 0) t--;
  }
  int top() {
    if (t >= 0) return arr[t];
    return -1;
  }
  bool empty() {
    if (t == -1)
      return true;
    else
      return false;
  }
};
const int mod = 1e9 + 7;
const int mxN = 1e5 + 5;
const string ln = "\n";
long long lpow(long long a, long long x, long long M = mod) {
  if (a == 0) return 0;
  if (x == 0) return (a % M + M) % M;
  long long res = 1;
  a = (a % M + M) % M;
  while (x > 0) {
    if (x & 1) res = (res * a) % M;
    a = (a * a) % M;
    x = x >> 1;
  }
  return res;
}
long long extendedEuler(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = extendedEuler(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long modInverse(long long a, long long M = mod) {
  long long x, y;
  long long g = extendedEuler(a, M, &x, &y);
  if (g != 1)
    return -1;
  else
    return (x % M + M) % M;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = int(0); i <= int(n - 1); ++i) cin >> a[i];
  struct stak S;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    while (!S.empty() && a[i] > S.top()) S.pop();
    if (!S.empty()) ans = max(ans, S.top() ^ a[i]);
    S.push(a[i]);
  }
  while (!S.empty()) S.pop();
  for (int i = n - 1; i >= 0; i--) {
    while (!S.empty() && a[i] > S.top()) S.pop();
    if (!S.empty()) ans = max(ans, S.top() ^ a[i]);
    S.push(a[i]);
  }
  cout << ans << "\n";
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
}
