#include <bits/stdc++.h>
using namespace std;
int pre[100001][25];
int arr[100001];
int n;
int get_log(int n) {
  if (n == 1) return 0;
  int res = 0;
  while (n > (1 << res)) res++;
  res--;
  return res;
}
int get_max(int l, int r) {
  if (l > r) return -1;
  int k = get_log(r - l + 1);
  if (arr[pre[l][k]] >= arr[pre[r - (1 << k) + 1][k]]) return pre[l][k];
  return pre[r - (1 << k) + 1][k];
}
int solve(int l, int r) {
  if (l >= r) return -1;
  int mx = get_max(l, r);
  int ans = max(solve(l, mx - 1), solve(mx + 1, r));
  int mxr = get_max(mx + 1, r);
  int mxl = get_max(l, mx - 1);
  while (mxr != -1) {
    ans = max(ans, arr[mx] ^ arr[mxr]);
    mxr = get_max(mx + 1, mxr - 1);
  }
  while (mxl != -1) {
    ans = max(ans, arr[mx] ^ arr[mxl]);
    mxl = get_max(mxl + 1, mx - 1);
  }
  return ans;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    pre[i][0] = i;
  }
  arr[n] = -0x7fffffff;
  for (int j = 1; (1 << j) <= n; j++)
    for (int i = 0; i + (1 << j) - 1 < n; i++)
      if (arr[pre[i][j - 1]] > arr[pre[i + (1 << (j - 1))][j - 1]])
        pre[i][j] = pre[i][j - 1];
      else
        pre[i][j] = pre[i + (1 << (j - 1))][j - 1];
  cout << solve(0, n - 1) << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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 pre[100001][25];
int arr[100001];
int n;
int get_log(int n) {
  if (n == 1) return 0;
  int res = 0;
  while (n > (1 << res)) res++;
  res--;
  return res;
}
int get_max(int l, int r) {
  if (l > r) return -1;
  int k = get_log(r - l + 1);
  if (arr[pre[l][k]] >= arr[pre[r - (1 << k) + 1][k]]) return pre[l][k];
  return pre[r - (1 << k) + 1][k];
}
int solve(int l, int r) {
  if (l >= r) return -1;
  int mx = get_max(l, r);
  int ans = max(solve(l, mx - 1), solve(mx + 1, r));
  int mxr = get_max(mx + 1, r);
  int mxl = get_max(l, mx - 1);
  while (mxr != -1) {
    ans = max(ans, arr[mx] ^ arr[mxr]);
    mxr = get_max(mx + 1, mxr - 1);
  }
  while (mxl != -1) {
    ans = max(ans, arr[mx] ^ arr[mxl]);
    mxl = get_max(mxl + 1, mx - 1);
  }
  return ans;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    pre[i][0] = i;
  }
  arr[n] = -0x7fffffff;
  for (int j = 1; (1 << j) <= n; j++)
    for (int i = 0; i + (1 << j) - 1 < n; i++)
      if (arr[pre[i][j - 1]] > arr[pre[i + (1 << (j - 1))][j - 1]])
        pre[i][j] = pre[i][j - 1];
      else
        pre[i][j] = pre[i + (1 << (j - 1))][j - 1];
  cout << solve(0, n - 1) << endl;
  return 0;
}

```