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 a[100001];
int BIT[100001];
std::pair<int, int> d[100001];
void update(int x) {
  while (x <= 100000) {
    BIT[x]++;
    x += x & -x;
  }
}
int get(int x) {
  int r = 0;
  while (x) {
    r += BIT[x];
    x -= x & -x;
  }
  return r;
}
int main() {
  int max = 0;
  int l, r, mid;
  int i, n;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    d[i].first = a[i];
    d[i].second = i;
  }
  std::sort(d + 1, d + n + 1);
  for (i = n; i >= 1; i--) {
    if (d[i].second > 1) {
      l = 1;
      r = d[i].second - 1;
      while (l < r) {
        mid = (l + r + 1) / 2;
        if (get(d[i].second - 1) - get(mid - 1))
          l = mid;
        else
          r = mid - 1;
      }
      if (get(l) - get(l - 1) && (d[i].first ^ a[l]) > max)
        max = d[i].first ^ a[l];
    }
    if (d[i].second < n) {
      l = d[i].second + 1;
      r = n;
      while (l < r) {
        mid = (l + r) / 2;
        if (get(mid) - get(d[i].second))
          r = mid;
        else
          l = mid + 1;
      }
      if (get(r) - get(r - 1) && (d[i].first ^ a[r]) > max)
        max = d[i].first ^ a[r];
    }
    update(d[i].second);
  }
  printf("%d", max);
}
