#include <bits/stdc++.h>
using namespace std;
int main() {
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    int a[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    int f[n], b[n];
    f[0] = -1;
    stack<int> s;
    s.push(0);
    for (int i = 1; i < n; i++) {
      while (!s.empty() && a[s.top()] < a[i]) s.pop();
      if (!s.empty())
        f[i] = s.top();
      else {
        f[i] = -1;
      }
      s.push(i);
    }
    stack<int> st;
    b[n - 1] = -1;
    st.push(n - 1);
    for (int i = n - 2; i >= 0; i--) {
      while (!st.empty() && a[st.top()] < a[i]) st.pop();
      if (!st.empty())
        b[i] = st.top();
      else {
        b[i] = -1;
      }
      st.push(i);
    }
    int ans = INT_MIN;
    for (int i = 0; i < n; i++) {
      if (f[i] != -1) ans = max(ans, a[i] ^ a[f[i]]);
      if (b[i] != -1) ans = max(ans, a[i] ^ a[b[i]]);
    }
    cout << ans << "\n";
  }
}

### Prompt

Your challenge is to write 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 main() {
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    int a[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    int f[n], b[n];
    f[0] = -1;
    stack<int> s;
    s.push(0);
    for (int i = 1; i < n; i++) {
      while (!s.empty() && a[s.top()] < a[i]) s.pop();
      if (!s.empty())
        f[i] = s.top();
      else {
        f[i] = -1;
      }
      s.push(i);
    }
    stack<int> st;
    b[n - 1] = -1;
    st.push(n - 1);
    for (int i = n - 2; i >= 0; i--) {
      while (!st.empty() && a[st.top()] < a[i]) st.pop();
      if (!st.empty())
        b[i] = st.top();
      else {
        b[i] = -1;
      }
      st.push(i);
    }
    int ans = INT_MIN;
    for (int i = 0; i < n; i++) {
      if (f[i] != -1) ans = max(ans, a[i] ^ a[f[i]]);
      if (b[i] != -1) ans = max(ans, a[i] ^ a[b[i]]);
    }
    cout << ans << "\n";
  }
}

```