#include <bits/stdc++.h>
int ary[100005], N, stk[100005], sn, ans;
void check(int x, int y) {
  if (ans < (x ^ y)) ans = x ^ y;
}
int main() {
  int i;
  scanf("%d", &N);
  for (i = 0; i < N; i++) scanf("%d", ary + i);
  for (i = 0; i < N; i++) {
    if (sn > 0) check(stk[sn - 1], ary[i]);
    while (sn > 0 && stk[sn - 1] <= ary[i]) check(stk[--sn], ary[i]);
    if (sn > 0) check(stk[sn - 1], ary[i]);
    if (sn == 0 || ary[i] < stk[sn - 1]) stk[sn++] = ary[i];
  }
  printf("%d\n", ans);
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
int ary[100005], N, stk[100005], sn, ans;
void check(int x, int y) {
  if (ans < (x ^ y)) ans = x ^ y;
}
int main() {
  int i;
  scanf("%d", &N);
  for (i = 0; i < N; i++) scanf("%d", ary + i);
  for (i = 0; i < N; i++) {
    if (sn > 0) check(stk[sn - 1], ary[i]);
    while (sn > 0 && stk[sn - 1] <= ary[i]) check(stk[--sn], ary[i]);
    if (sn > 0) check(stk[sn - 1], ary[i]);
    if (sn == 0 || ary[i] < stk[sn - 1]) stk[sn++] = ary[i];
  }
  printf("%d\n", ans);
}

```