#include <bits/stdc++.h>
using namespace std;
mt19937_64 random_num(
    chrono::high_resolution_clock::now().time_since_epoch().count());
const int MOD = 1000000007;
const int inf = 2000000000;
const int mod = 1007050321;
const double e = 0.000001;
const int N = 200005;
const bool aut = 788480;
const int base = 521;
int n, a[N];
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  deque<int> dq;
  a[0] = inf;
  dq.push_back(0);
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    while (dq.size() && a[dq.front()] <= a[i]) dq.pop_front();
    if (dq.front() != 0) ans = max(ans, a[i] ^ a[dq.front()]);
    dq.push_front(i);
  }
  dq.clear();
  a[n + 1] = inf;
  dq.push_back(n + 1);
  for (int i = n; i >= 1; --i) {
    while (dq.front() && a[dq.front()] <= a[i]) dq.pop_front();
    if (dq.front() != n + 1) ans = max(ans, a[dq.front()] ^ a[i]);
    dq.push_front(i);
  }
  cout << ans;
  return 0;
}

### Prompt

Please formulate 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;
mt19937_64 random_num(
    chrono::high_resolution_clock::now().time_since_epoch().count());
const int MOD = 1000000007;
const int inf = 2000000000;
const int mod = 1007050321;
const double e = 0.000001;
const int N = 200005;
const bool aut = 788480;
const int base = 521;
int n, a[N];
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  deque<int> dq;
  a[0] = inf;
  dq.push_back(0);
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    while (dq.size() && a[dq.front()] <= a[i]) dq.pop_front();
    if (dq.front() != 0) ans = max(ans, a[i] ^ a[dq.front()]);
    dq.push_front(i);
  }
  dq.clear();
  a[n + 1] = inf;
  dq.push_back(n + 1);
  for (int i = n; i >= 1; --i) {
    while (dq.front() && a[dq.front()] <= a[i]) dq.pop_front();
    if (dq.front() != n + 1) ans = max(ans, a[dq.front()] ^ a[i]);
    dq.push_front(i);
  }
  cout << ans;
  return 0;
}

```