#include <bits/stdc++.h>
int a[100005], n, max, bi[50], k, i, j, ans, p;
bool bj;
template <class T>
void up(T& a, T b) {
  if (a < b) a = b;
}
template <class T>
void down(T& a, T b) {
  if (a > b) a = b;
}
void work1(int l, int r) {
  if (l == 0) return;
  int max = 0;
  for (int i = l + 1; i <= r; i++) up(max, a[i]), up(ans, a[l] ^ max);
}
void work2(int l, int r) {
  int max = 0;
  for (int i = r - 1; i >= l; i--) up(max, a[i]), up(ans, a[r] ^ max);
}
int main() {
  scanf("%d", &n);
  for (bi[0] = i = 1; i <= 31; i++) bi[i] = bi[i - 1] << 1;
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), up(max, a[i]);
  for (k = 30; (max & bi[k]) == 0; k--)
    ;
  for (; !bj && k >= 0; k--) {
    p = 0;
    for (i = 1; i <= n; i++)
      if ((a[i] & bi[k]) != 0) {
        if (p != i - 1) work1(p, i - 1), work2(p + 1, i), bj = 1;
        p = i;
      }
    if (p != n && p != 0) work1(p, n), bj = 1;
  }
  printf("%d", ans);
  return 0;
}

### Prompt

Your task is to create 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>
int a[100005], n, max, bi[50], k, i, j, ans, p;
bool bj;
template <class T>
void up(T& a, T b) {
  if (a < b) a = b;
}
template <class T>
void down(T& a, T b) {
  if (a > b) a = b;
}
void work1(int l, int r) {
  if (l == 0) return;
  int max = 0;
  for (int i = l + 1; i <= r; i++) up(max, a[i]), up(ans, a[l] ^ max);
}
void work2(int l, int r) {
  int max = 0;
  for (int i = r - 1; i >= l; i--) up(max, a[i]), up(ans, a[r] ^ max);
}
int main() {
  scanf("%d", &n);
  for (bi[0] = i = 1; i <= 31; i++) bi[i] = bi[i - 1] << 1;
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), up(max, a[i]);
  for (k = 30; (max & bi[k]) == 0; k--)
    ;
  for (; !bj && k >= 0; k--) {
    p = 0;
    for (i = 1; i <= n; i++)
      if ((a[i] & bi[k]) != 0) {
        if (p != i - 1) work1(p, i - 1), work2(p + 1, i), bj = 1;
        p = i;
      }
    if (p != n && p != 0) work1(p, n), bj = 1;
  }
  printf("%d", ans);
  return 0;
}

```