#include <bits/stdc++.h>
using namespace std;
int d[2 * 140000], p, a[140000], n;
int next(int i) {
  int j, l = a[i];
  for (i++, j = p - 1, i += p, j += p; i <= j && d[i] <= l;
       i += i & 1, i >>= 1, j >>= 1)
    ;
  if (i > j) return p;
  for (; i < p; i <<= 1, i += d[i] <= l)
    ;
  return i - p;
}
int sol() {
  int i, j, k, r;
  for (p = 1; p < n; p *= 2)
    ;
  for (i = 0; i < n; d[p + i] = a[i], i++)
    ;
  for (i = p - 1; i > 0; d[i] = max(d[2 * i], d[2 * i + 1]), i--)
    ;
  for (r = 0, i = 0; i + 1 < n; i++)
    for (j = min(next(i), n), k = i + 1; k < j;
         r = max(r, a[i] ^ a[k]), k = next(k))
      ;
  return r;
}
int main() {
  int i, r;
  for (scanf("%d", &n), i = 0; i < n; scanf("%d", &a[i]), i++)
    ;
  r = sol();
  reverse(a, a + n);
  r = max(r, sol());
  printf("%d\n", r);
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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}.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int d[2 * 140000], p, a[140000], n;
int next(int i) {
  int j, l = a[i];
  for (i++, j = p - 1, i += p, j += p; i <= j && d[i] <= l;
       i += i & 1, i >>= 1, j >>= 1)
    ;
  if (i > j) return p;
  for (; i < p; i <<= 1, i += d[i] <= l)
    ;
  return i - p;
}
int sol() {
  int i, j, k, r;
  for (p = 1; p < n; p *= 2)
    ;
  for (i = 0; i < n; d[p + i] = a[i], i++)
    ;
  for (i = p - 1; i > 0; d[i] = max(d[2 * i], d[2 * i + 1]), i--)
    ;
  for (r = 0, i = 0; i + 1 < n; i++)
    for (j = min(next(i), n), k = i + 1; k < j;
         r = max(r, a[i] ^ a[k]), k = next(k))
      ;
  return r;
}
int main() {
  int i, r;
  for (scanf("%d", &n), i = 0; i < n; scanf("%d", &a[i]), i++)
    ;
  r = sol();
  reverse(a, a + n);
  r = max(r, sol());
  printf("%d\n", r);
  return 0;
}

```