#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    long long int n;
    cin >> n;
    vector<long long int> a(n);
    for (int i = 0; i < n; ++i) cin >> a[i];
    vector<long long int> st;
    long long int ans = 0;
    for (int i = 0; i < n; ++i) {
      while (!st.empty() && st.back() < a[i]) {
        st.pop_back();
      }
      st.push_back(a[i]);
      if ((int)st.size() >= 2) {
        ans = max(ans, st[(int)st.size() - 1] ^ st[(int)st.size() - 2]);
      }
    }
    for (int i = n - 1; i >= 0; i--) {
      while (!st.empty() && st.back() < a[i]) {
        st.pop_back();
      }
      st.push_back(a[i]);
      if ((int)st.size() >= 2) {
        ans = max(ans, st[(int)st.size() - 1] ^ st[(int)st.size() - 2]);
      }
    }
    cout << ans << "\n";
  }
  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;
const long long int MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    long long int n;
    cin >> n;
    vector<long long int> a(n);
    for (int i = 0; i < n; ++i) cin >> a[i];
    vector<long long int> st;
    long long int ans = 0;
    for (int i = 0; i < n; ++i) {
      while (!st.empty() && st.back() < a[i]) {
        st.pop_back();
      }
      st.push_back(a[i]);
      if ((int)st.size() >= 2) {
        ans = max(ans, st[(int)st.size() - 1] ^ st[(int)st.size() - 2]);
      }
    }
    for (int i = n - 1; i >= 0; i--) {
      while (!st.empty() && st.back() < a[i]) {
        st.pop_back();
      }
      st.push_back(a[i]);
      if ((int)st.size() >= 2) {
        ans = max(ans, st[(int)st.size() - 1] ^ st[(int)st.size() - 2]);
      }
    }
    cout << ans << "\n";
  }
  return 0;
}

```