#include <bits/stdc++.h>
using namespace std;
long long pow(long long b, long long e, long long m) {
  long long t = 1;
  for (; e; e >>= 1, b = b * b % m) e & 1 ? t = t * b % m : 0;
  return t;
}
template <class T>
inline bool chkmin(T &a, T b) {
  return a > b ? a = b, true : false;
}
template <class T>
inline bool chkmax(T &a, T b) {
  return a < b ? a = b, true : false;
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <typename T>
T gcd(T x, T y) {
  for (T t; x; t = x, x = y % x, y = t)
    ;
  return y;
}
template <class edge>
struct Graph {
  vector<vector<edge> > adj;
  Graph(int n) {
    adj.clear();
    adj.resize(n + 5);
  }
  Graph() { adj.clear(); }
  void resize(int n) { adj.resize(n + 5); }
  void add(int s, edge e) { adj[s].push_back(e); }
  void del(int s, edge e) { adj[s].erase(find(iter(adj[s]), e)); }
  vector<edge> &operator[](int t) { return adj[t]; }
};
int main() {
  ios_base::sync_with_stdio(false);
  int n, ans = 0;
  cin >> n;
  stack<int> Q;
  for (int i = 1, v; i <= n; ++i) {
    cin >> v;
    for (; !Q.empty(); Q.pop()) {
      int x = Q.top();
      chkmax(ans, v ^ x);
      if (x > v) {
        break;
      }
    }
    Q.push(v);
  }
  cout << setprecision(15) << ans << endl;
  return 0;
}

### Prompt

Your task is to create 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;
long long pow(long long b, long long e, long long m) {
  long long t = 1;
  for (; e; e >>= 1, b = b * b % m) e & 1 ? t = t * b % m : 0;
  return t;
}
template <class T>
inline bool chkmin(T &a, T b) {
  return a > b ? a = b, true : false;
}
template <class T>
inline bool chkmax(T &a, T b) {
  return a < b ? a = b, true : false;
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <typename T>
T gcd(T x, T y) {
  for (T t; x; t = x, x = y % x, y = t)
    ;
  return y;
}
template <class edge>
struct Graph {
  vector<vector<edge> > adj;
  Graph(int n) {
    adj.clear();
    adj.resize(n + 5);
  }
  Graph() { adj.clear(); }
  void resize(int n) { adj.resize(n + 5); }
  void add(int s, edge e) { adj[s].push_back(e); }
  void del(int s, edge e) { adj[s].erase(find(iter(adj[s]), e)); }
  vector<edge> &operator[](int t) { return adj[t]; }
};
int main() {
  ios_base::sync_with_stdio(false);
  int n, ans = 0;
  cin >> n;
  stack<int> Q;
  for (int i = 1, v; i <= n; ++i) {
    cin >> v;
    for (; !Q.empty(); Q.pop()) {
      int x = Q.top();
      chkmax(ans, v ^ x);
      if (x > v) {
        break;
      }
    }
    Q.push(v);
  }
  cout << setprecision(15) << ans << endl;
  return 0;
}

```