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 MAXN = 100010;
int n, big, ans, a[MAXN], b[MAXN];
bool flag;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  do {
    flag = false;
    big = 0;
    int tmp = 0;
    for (int i = 0; i <= 30; ++i)
      for (int j = 1; j <= n; ++j)
        if ((1 << i) & a[j]) big = (1 << i);
    for (int i = 1; i <= n; ++i)
      if (a[i] & big) ++tmp;
    if (tmp == n) {
      for (int i = 1; i <= n; ++i) a[i] -= big;
      flag = true;
    }
  } while (flag);
  b[++b[0]] = 0;
  for (int i = 1; i <= n; ++i)
    if (a[i] & big) b[++b[0]] = i;
  b[++b[0]] = n + 1;
  for (int i = 2; i < b[0]; ++i) {
    int tmp = 0;
    for (int j = b[i] - 1; j > b[i - 1]; --j) {
      tmp = max(tmp, a[j]);
      ans = max(ans, tmp ^ a[b[i]]);
    }
    tmp = 0;
    for (int j = b[i] + 1; j < b[i + 1]; ++j) {
      tmp = max(tmp, a[j]);
      ans = max(ans, tmp ^ a[b[i]]);
    }
  }
  printf("%d\n", ans);
  return 0;
}
