#include <bits/stdc++.h>
using namespace std;
struct Node {
  int num, no;
} s[101010];
map<int, int> mp1, mp2;
bool cmp(Node a, Node b) { return a.num > b.num; }
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &s[i].num);
    s[i].no = i;
  }
  int ans = 0;
  mp1.clear();
  mp2.clear();
  mp1[0] = mp2[n + 1] = s[1].num;
  mp1[n + 1] = mp2[0] = s[n].num;
  sort(s + 1, s + 1 + n, cmp);
  for (int i = 1; i <= n; i++) {
    mp1[s[i].no] = s[i].num;
    mp2[n - s[i].no + 1] = s[i].num;
    map<int, int>::iterator it = mp1.upper_bound(s[i].no);
    if (it->first != n + 1) ans = max(ans, it->second ^ s[i].num);
    it = mp2.upper_bound(n - s[i].no + 1);
    if (it->first != n + 1) ans = max(ans, it->second ^ s[i].num);
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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;
struct Node {
  int num, no;
} s[101010];
map<int, int> mp1, mp2;
bool cmp(Node a, Node b) { return a.num > b.num; }
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &s[i].num);
    s[i].no = i;
  }
  int ans = 0;
  mp1.clear();
  mp2.clear();
  mp1[0] = mp2[n + 1] = s[1].num;
  mp1[n + 1] = mp2[0] = s[n].num;
  sort(s + 1, s + 1 + n, cmp);
  for (int i = 1; i <= n; i++) {
    mp1[s[i].no] = s[i].num;
    mp2[n - s[i].no + 1] = s[i].num;
    map<int, int>::iterator it = mp1.upper_bound(s[i].no);
    if (it->first != n + 1) ans = max(ans, it->second ^ s[i].num);
    it = mp2.upper_bound(n - s[i].no + 1);
    if (it->first != n + 1) ans = max(ans, it->second ^ s[i].num);
  }
  printf("%d\n", ans);
  return 0;
}

```