#include <bits/stdc++.h>
using namespace std;
const int N = 100100;
int a[N * 32], child[N * 32][3], cnt, p[N], ans;
vector<int> vc;
void add(int val, int sgn) {
  int loc(0);
  a[0] += sgn;
  for (int i = 30; i >= 0; i--) {
    bool t = (((1 << i) & val) > 0);
    if (!child[loc][t]) child[loc][t] = ++cnt;
    loc = child[loc][t];
    a[loc] += sgn;
  }
  return;
}
int get(int val) {
  int ans(0), loc(0), flag(0);
  for (int i = 30; i >= 0; i--) {
    bool t = 1 - (((1 << i) & val) > 0);
    if (child[loc][t] && a[child[loc][t]] && (flag || !t)) {
      ans += (1 << i);
      loc = child[loc][t];
      flag = 1;
    } else if (child[loc][1 - t])
      loc = child[loc][1 - t];
    else
      return 0;
  }
  return ans;
}
void idk(int i) {
  ans = max(ans, get(p[i]));
  while (!vc.empty() && vc.back() < p[i]) {
    add(vc.back(), -1);
    vc.pop_back();
  }
  vc.push_back(p[i]);
  add(p[i], 1);
  return;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p[i];
  add(p[0], 1);
  vc.push_back(p[0]);
  for (int i = 1; i < n; i++) idk(i);
  while (vc.size()) {
    add(vc.back(), -1);
    vc.pop_back();
  }
  add(p[n - 1], 1);
  vc.push_back(p[n - 1]);
  for (int i = n - 2; i >= 0; i--) idk(i);
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const int N = 100100;
int a[N * 32], child[N * 32][3], cnt, p[N], ans;
vector<int> vc;
void add(int val, int sgn) {
  int loc(0);
  a[0] += sgn;
  for (int i = 30; i >= 0; i--) {
    bool t = (((1 << i) & val) > 0);
    if (!child[loc][t]) child[loc][t] = ++cnt;
    loc = child[loc][t];
    a[loc] += sgn;
  }
  return;
}
int get(int val) {
  int ans(0), loc(0), flag(0);
  for (int i = 30; i >= 0; i--) {
    bool t = 1 - (((1 << i) & val) > 0);
    if (child[loc][t] && a[child[loc][t]] && (flag || !t)) {
      ans += (1 << i);
      loc = child[loc][t];
      flag = 1;
    } else if (child[loc][1 - t])
      loc = child[loc][1 - t];
    else
      return 0;
  }
  return ans;
}
void idk(int i) {
  ans = max(ans, get(p[i]));
  while (!vc.empty() && vc.back() < p[i]) {
    add(vc.back(), -1);
    vc.pop_back();
  }
  vc.push_back(p[i]);
  add(p[i], 1);
  return;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p[i];
  add(p[0], 1);
  vc.push_back(p[0]);
  for (int i = 1; i < n; i++) idk(i);
  while (vc.size()) {
    add(vc.back(), -1);
    vc.pop_back();
  }
  add(p[n - 1], 1);
  vc.push_back(p[n - 1]);
  for (int i = n - 2; i >= 0; i--) idk(i);
  cout << ans << endl;
  return 0;
}

```