#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 100;
stack<long long> a;
long long nextm[N];
long long nextn[N];
vector<long long> s;
void make(int n) {
  fill(nextn, nextn + N, 0);
  fill(nextm, nextm + N, 0);
  long long v;
  for (int i = 0; i < n; i++) {
    while (a.size() && s[a.top()] < s[i]) a.pop();
    nextn[i] = (a.size() ? a.top() : -1LL);
    a.push(i);
  }
  while (a.size()) a.pop();
  for (int i = n - 1; i > 0; i--) {
    while (a.size() && s[a.top()] < s[i]) a.pop();
    nextm[i] = (a.size() ? a.top() : -1LL);
    a.push(i);
  }
  while (a.size()) a.pop();
}
int main() {
  long long n, ans;
  cin >> n;
  s.resize(n);
  for (int i = 0; i < n; i++) cin >> s[i], nextm[i] = nextn[i] = -1;
  ans = s[0] ^ s[1];
  make(n);
  for (int i = 0; i < n; i++) {
    if (nextn[i] > -1) ans = max(ans, (s[i] ^ s[nextn[i]]));
    if (nextm[i] > -1) ans = max(ans, (s[i] ^ s[nextm[i]]));
  }
  cout << ans << endl;
}

### Prompt

Develop a solution in CPP 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 = 1e5 + 100;
stack<long long> a;
long long nextm[N];
long long nextn[N];
vector<long long> s;
void make(int n) {
  fill(nextn, nextn + N, 0);
  fill(nextm, nextm + N, 0);
  long long v;
  for (int i = 0; i < n; i++) {
    while (a.size() && s[a.top()] < s[i]) a.pop();
    nextn[i] = (a.size() ? a.top() : -1LL);
    a.push(i);
  }
  while (a.size()) a.pop();
  for (int i = n - 1; i > 0; i--) {
    while (a.size() && s[a.top()] < s[i]) a.pop();
    nextm[i] = (a.size() ? a.top() : -1LL);
    a.push(i);
  }
  while (a.size()) a.pop();
}
int main() {
  long long n, ans;
  cin >> n;
  s.resize(n);
  for (int i = 0; i < n; i++) cin >> s[i], nextm[i] = nextn[i] = -1;
  ans = s[0] ^ s[1];
  make(n);
  for (int i = 0; i < n; i++) {
    if (nextn[i] > -1) ans = max(ans, (s[i] ^ s[nextn[i]]));
    if (nextm[i] > -1) ans = max(ans, (s[i] ^ s[nextm[i]]));
  }
  cout << ans << endl;
}

```