#include <bits/stdc++.h>
using namespace std;
long kt(long a, long b) {
  long dem1, dem2;
  dem1 = 0;
  while (a) {
    if (a % 2) dem1++;
    a /= 2;
  }
  dem2 = 0;
  while (b) {
    if (b % 2) dem2++;
    b /= 2;
  }
  if (dem1 > dem2) return 1;
  return 0;
}
int main() {
  long a[100009];
  long n, top, u, t, res, i;
  top = 0;
  res = 0;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> u;
    while ((top) && (a[top] < u)) {
      t = a[top] ^ u;
      if (t > res) res = t;
      top--;
    }
    top++;
    a[top] = u;
    if (top != 1) {
      t = a[top - 1] ^ u;
      if (t > res) res = t;
    }
  }
  cout << res;
  return 0;
}

### Prompt

Please formulate a Cpp solution to 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;
long kt(long a, long b) {
  long dem1, dem2;
  dem1 = 0;
  while (a) {
    if (a % 2) dem1++;
    a /= 2;
  }
  dem2 = 0;
  while (b) {
    if (b % 2) dem2++;
    b /= 2;
  }
  if (dem1 > dem2) return 1;
  return 0;
}
int main() {
  long a[100009];
  long n, top, u, t, res, i;
  top = 0;
  res = 0;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> u;
    while ((top) && (a[top] < u)) {
      t = a[top] ^ u;
      if (t > res) res = t;
      top--;
    }
    top++;
    a[top] = u;
    if (top != 1) {
      t = a[top - 1] ^ u;
      if (t > res) res = t;
    }
  }
  cout << res;
  return 0;
}

```