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;
int64_t power(int64_t a, int64_t b) {
  int64_t ans = 1;
  while (b) {
    if (b & 1) {
      ans = (ans * a) % (int64_t)(1e9 + 7);
    }
    a = (a * a) % (int64_t)(1e9 + 7);
    b >>= 1;
  }
  return ans;
}
int64_t mmi(int64_t n) { return power(n, (int64_t)(1e9 + 7) - 2); }
int64_t two(int64_t n) { return (1LL << n); }
void onbit(int64_t &n, int64_t x) { n |= two(x); }
void offbit(int64_t &n, int64_t x) { n &= ~two(x); }
int64_t cntbit(int64_t n) {
  int64_t res = 0;
  while (n and ++res) n -= n & -n;
  return res;
}
int64_t dx[] = {-2, -2, +1, -1};
int64_t dy[] = {+1, -1, -2, -2};
int n, a[(int64_t)(1e5 + 5)], l[(int64_t)(1e5 + 5)], r[(int64_t)(1e5 + 5)];
stack<int> st;
void solve() {
  int i;
  cin >> n;
  for (i = 1; i < n + 1; i++) cin >> a[i];
  int ans = 0;
  for (i = n; i > 0; i--) {
    while (!st.empty() and a[st.top()] < a[i]) {
      st.pop();
    }
    if (!st.empty()) ans = max(ans, a[i] ^ a[st.top()]);
    st.push(i);
  }
  while (!st.empty()) {
    int64_t x = st.top();
    st.pop();
    if (st.empty()) break;
    ans = max(ans, a[x] ^ a[st.top()]);
  }
  for (i = 1; i < n + 1; i++) {
    while (!st.empty() and a[st.top()] < a[i]) {
      st.pop();
    }
    if (!st.empty()) ans = max(ans, a[i] ^ a[st.top()]);
    st.push(i);
  }
  while (!st.empty()) {
    int64_t x = st.top();
    st.pop();
    if (st.empty()) break;
    ans = max(ans, a[x] ^ a[st.top()]);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int64_t t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
