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 mod = 1e9 + 7;
const int oo = INT_MAX;
const int N = 1e6 + 1;
long long n, a[N], d[N];
inline void nhap() {
  d[0] = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    d[i] = max(d[i - 1], a[i]);
  }
}
stack<int> st;
inline void solve() {
  st.push(1);
  long long ans = -1;
  for (int i = 2; i <= n; i++) {
    while (st.size() != 0 && a[st.top()] < a[i]) {
      st.pop();
    }
    if (st.empty())
      ans = max(ans, a[i] ^ d[i - 1]);
    else
      ans = max(ans, a[st.top()] ^ a[i]);
    st.push(i);
  }
  while (st.empty() == false) st.pop();
  d[n] = a[n];
  for (int i = n - 1; i >= 1; i--) d[i] = max(d[i + 1], a[i]);
  st.push(n);
  for (int i = n - 1; i >= 1; i--) {
    while (st.size() != 0 && a[st.top()] < a[i]) st.pop();
    if (st.empty())
      ans = max(ans, a[i] ^ d[i + 1]);
    else
      ans = max(ans, a[i] ^ a[st.top()]);
    st.push(i);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    nhap();
    solve();
  }
}
