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;
int i, j, k, l, cur, a[100005], n, ans;
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (j = 31; j >= 0; j--) {
    cur = 0;
    for (i = 1; i <= n; i++)
      if (a[i] & (1 << j)) cur++;
    if (cur == n || cur == 0) continue;
    l = 1;
    for (i = 1; i <= n; i++)
      if (a[i] & (1 << j)) {
        cur = 0;
        for (k = i - 1; k >= l; k--) {
          cur = (cur > a[k] ? cur : a[k]);
          ans = (ans > (a[i] ^ cur) ? ans : (a[i] ^ cur));
        }
        l = i + 1;
      }
    reverse(a + 1, a + n + 1);
    l = 1;
    for (i = 1; i <= n; i++)
      if (a[i] & (1 << j)) {
        cur = 0;
        for (k = i - 1; k >= l; k--) {
          cur = (cur > a[k] ? cur : a[k]);
          ans = (ans > (a[i] ^ cur) ? ans : (a[i] ^ cur));
        }
        l = i + 1;
      }
    reverse(a + 1, a + n + 1);
    break;
  }
  cout << ans << endl;
}
