#include <bits/stdc++.h>
using namespace std;
int stmp;
const int MAX = 100000;
const int INF = 1000000001;
int arr[MAX];
int n;
int main(int argc, char *argv[]) {
  stmp = scanf("%d", &n);
  for (int i = 0; i < (n); ++i) stmp = scanf("%d", arr + i);
  int res = 0;
  int b = 30;
  while (1) {
    int cnt = 0;
    for (int i = 0; i < (n); ++i)
      if (arr[i] & (1 << b)) ++cnt;
    if (cnt != 0 && cnt != n) break;
    --b;
  }
  for (int i = 0; i < (n); ++i)
    if (arr[i] & (1 << b)) {
      vector<int> v;
      v.push_back(arr[i]);
      for (int j = i + 1; j < n && !(arr[j] & (1 << b)); ++j) {
        v.push_back(arr[j]);
        sort((v).begin(), (v).end(), greater<int>());
        if (((int)(v).size()) > 2) v.pop_back();
        res = max(res, v[0] ^ v[1]);
      }
      v.clear();
      v.push_back(arr[i]);
      for (int j = i - 1; j >= 0 && !(arr[j] & (1 << b)); --j) {
        v.push_back(arr[j]);
        sort((v).begin(), (v).end(), greater<int>());
        if (((int)(v).size()) > 2) v.pop_back();
        res = max(res, v[0] ^ v[1]);
      }
    }
  printf("%d\n", res);
  return 0;
}

### Prompt

Generate 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;
int stmp;
const int MAX = 100000;
const int INF = 1000000001;
int arr[MAX];
int n;
int main(int argc, char *argv[]) {
  stmp = scanf("%d", &n);
  for (int i = 0; i < (n); ++i) stmp = scanf("%d", arr + i);
  int res = 0;
  int b = 30;
  while (1) {
    int cnt = 0;
    for (int i = 0; i < (n); ++i)
      if (arr[i] & (1 << b)) ++cnt;
    if (cnt != 0 && cnt != n) break;
    --b;
  }
  for (int i = 0; i < (n); ++i)
    if (arr[i] & (1 << b)) {
      vector<int> v;
      v.push_back(arr[i]);
      for (int j = i + 1; j < n && !(arr[j] & (1 << b)); ++j) {
        v.push_back(arr[j]);
        sort((v).begin(), (v).end(), greater<int>());
        if (((int)(v).size()) > 2) v.pop_back();
        res = max(res, v[0] ^ v[1]);
      }
      v.clear();
      v.push_back(arr[i]);
      for (int j = i - 1; j >= 0 && !(arr[j] & (1 << b)); --j) {
        v.push_back(arr[j]);
        sort((v).begin(), (v).end(), greater<int>());
        if (((int)(v).size()) > 2) v.pop_back();
        res = max(res, v[0] ^ v[1]);
      }
    }
  printf("%d\n", res);
  return 0;
}

```