#include <bits/stdc++.h>
using namespace std;
int s[2000000];
int n;
int sor = 0;
vector<int> cc;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", s + i);
  for (int i = 0; i < n; i++) sor |= s[i];
  int mm = 1;
  for (; mm <= sor; mm <<= 1)
    ;
  mm >>= 1;
  for (int i = 0; i < n; i++)
    if ((mm & s[i]) != 0) cc.push_back(i);
  while (cc.size() == n || cc.size() == 0) {
    cc.clear();
    mm >>= 1;
    for (int i = 0; i < n; i++)
      if ((mm & s[i]) != 0) cc.push_back(i);
  }
  int ii = 0;
  int imax = 0;
  int max2 = 0;
  for (int i = cc[0] - 1; i >= 0; i--) {
    max2 = max(s[i], max2);
    imax = max(imax, max2 ^ s[cc[0]]);
  }
  for (int i = 0; i < cc.size() - 1; i++) {
    max2 = 0;
    for (int j = cc[i] + 1; j <= cc[i + 1]; j++) {
      max2 = max(max2, s[j]);
      imax = max(imax, max2 ^ s[cc[i]]);
    }
    max2 = 0;
    for (int j = cc[i + 1] - 1; j > cc[i]; j--) {
      max2 = max(max2, s[j]);
      imax = max(imax, max2 ^ s[cc[i + 1]]);
    }
  }
  max2 = 0;
  for (int i = cc[cc.size() - 1] + 1; i < n; i++) {
    max2 = max(max2, s[i]);
    imax = max(imax, max2 ^ s[cc[cc.size() - 1]]);
  }
  printf("%d", imax);
}

### 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;
int s[2000000];
int n;
int sor = 0;
vector<int> cc;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", s + i);
  for (int i = 0; i < n; i++) sor |= s[i];
  int mm = 1;
  for (; mm <= sor; mm <<= 1)
    ;
  mm >>= 1;
  for (int i = 0; i < n; i++)
    if ((mm & s[i]) != 0) cc.push_back(i);
  while (cc.size() == n || cc.size() == 0) {
    cc.clear();
    mm >>= 1;
    for (int i = 0; i < n; i++)
      if ((mm & s[i]) != 0) cc.push_back(i);
  }
  int ii = 0;
  int imax = 0;
  int max2 = 0;
  for (int i = cc[0] - 1; i >= 0; i--) {
    max2 = max(s[i], max2);
    imax = max(imax, max2 ^ s[cc[0]]);
  }
  for (int i = 0; i < cc.size() - 1; i++) {
    max2 = 0;
    for (int j = cc[i] + 1; j <= cc[i + 1]; j++) {
      max2 = max(max2, s[j]);
      imax = max(imax, max2 ^ s[cc[i]]);
    }
    max2 = 0;
    for (int j = cc[i + 1] - 1; j > cc[i]; j--) {
      max2 = max(max2, s[j]);
      imax = max(imax, max2 ^ s[cc[i + 1]]);
    }
  }
  max2 = 0;
  for (int i = cc[cc.size() - 1] + 1; i < n; i++) {
    max2 = max(max2, s[i]);
    imax = max(imax, max2 ^ s[cc[cc.size() - 1]]);
  }
  printf("%d", imax);
}

```