#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC push_options
#pragma GCC optimize("unroll-loops")
using namespace std;
void doin() {
  cin.tie();
  cout.tie();
  ios::sync_with_stdio(0);
}
template <typename X, typename Y>
istream& operator>>(istream& in, pair<X, Y>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
void getv(T& i) {
  cin >> i;
}
template <typename T, typename... Ns>
void getv(vector<T>& v, int n, Ns... ns) {
  v.resize(n);
  for (auto& i : v) getv(i, ns...);
}
template <typename T>
void getv1(T& i) {
  cin >> i;
}
template <typename T, typename... Ns>
void getv1(vector<T>& v, int n, Ns... ns) {
  v.resize(n + 1);
  for (int i = 1; i <= n; i++) getv(v[i], ns...);
}
inline void getch(char& x) {
  while (x = getchar_unlocked(), x < 33) {
    ;
  }
}
inline void getstr(string& str) {
  str.clear();
  char cur;
  while (cur = getchar_unlocked(), cur < 33) {
    ;
  }
  while (cur > 32) {
    str += cur;
    cur = getchar_unlocked();
  }
}
template <typename T>
inline bool sc(T& num) {
  bool neg = 0;
  int c;
  num = 0;
  while (c = getchar_unlocked(), c < 33) {
    if (c == EOF) return false;
  }
  if (c == '-') {
    neg = 1;
    c = getchar_unlocked();
  }
  for (; c > 47; c = getchar_unlocked()) num = num * 10 + c - 48;
  if (neg) num *= -1;
  return true;
}
using AntonTsypko = void;
using arsijo = AntonTsypko;
using god = arsijo;
uniform_real_distribution<double> double_dist(0, 1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  doin();
  int n, mx, t;
  mx = 0;
  cin >> n;
  stack<int> mono;
  for (int i = 0; i < n; i++) {
    cin >> t;
    while (true) {
      if (!mono.empty()) mx = max(mono.top() ^ t, mx);
      if (!mono.empty() && mono.top() < t)
        mono.pop();
      else
        break;
    }
    mono.push(t);
  }
  cout << mx;
}

### Prompt

Please create a solution in cpp 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>
#pragma GCC optimize("Ofast")
#pragma GCC push_options
#pragma GCC optimize("unroll-loops")
using namespace std;
void doin() {
  cin.tie();
  cout.tie();
  ios::sync_with_stdio(0);
}
template <typename X, typename Y>
istream& operator>>(istream& in, pair<X, Y>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
void getv(T& i) {
  cin >> i;
}
template <typename T, typename... Ns>
void getv(vector<T>& v, int n, Ns... ns) {
  v.resize(n);
  for (auto& i : v) getv(i, ns...);
}
template <typename T>
void getv1(T& i) {
  cin >> i;
}
template <typename T, typename... Ns>
void getv1(vector<T>& v, int n, Ns... ns) {
  v.resize(n + 1);
  for (int i = 1; i <= n; i++) getv(v[i], ns...);
}
inline void getch(char& x) {
  while (x = getchar_unlocked(), x < 33) {
    ;
  }
}
inline void getstr(string& str) {
  str.clear();
  char cur;
  while (cur = getchar_unlocked(), cur < 33) {
    ;
  }
  while (cur > 32) {
    str += cur;
    cur = getchar_unlocked();
  }
}
template <typename T>
inline bool sc(T& num) {
  bool neg = 0;
  int c;
  num = 0;
  while (c = getchar_unlocked(), c < 33) {
    if (c == EOF) return false;
  }
  if (c == '-') {
    neg = 1;
    c = getchar_unlocked();
  }
  for (; c > 47; c = getchar_unlocked()) num = num * 10 + c - 48;
  if (neg) num *= -1;
  return true;
}
using AntonTsypko = void;
using arsijo = AntonTsypko;
using god = arsijo;
uniform_real_distribution<double> double_dist(0, 1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  doin();
  int n, mx, t;
  mx = 0;
  cin >> n;
  stack<int> mono;
  for (int i = 0; i < n; i++) {
    cin >> t;
    while (true) {
      if (!mono.empty()) mx = max(mono.top() ^ t, mx);
      if (!mono.empty() && mono.top() < t)
        mono.pop();
      else
        break;
    }
    mono.push(t);
  }
  cout << mx;
}

```