#include <bits/stdc++.h>
using namespace std;
void setIO(const string &name = "") {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  if (name.length()) {
    freopen((name + ".in").c_str(), "r", stdin);
    freopen((name + ".out").c_str(), "w", stdout);
  }
}
template <typename T>
void read(vector<T> &a) {
  for (auto &x : a) cin >> x;
}
template <typename T>
void read(vector<T> &a, long long n) {
  a.resize(n);
  for (auto &x : a) cin >> x;
}
template <typename T>
void print(vector<T> &a) {
  for (const auto &x : a) cout << x << ' ';
  cout << '\n';
}
void MOD(long long &x) {
  x %= 1000000007;
  if (x < 0) x += 1000000007;
}
long long gcd(long long a, long long b) { return !a ? b : gcd(b % a, a); }
long long solve(const vector<long long> &v) {
  long long res = v[0] xor v[1];
  long long n = (long long)(v).size();
  vector<long long> st;
  for (auto(i) = 0; i < n; i++) {
    while (!st.empty() and st.back() < v[i]) st.pop_back();
    st.emplace_back(v[i]);
    if ((long long)(st).size() >= 2)
      res = max(res, st[(long long)(st).size() - 1] xor
                         st[(long long)(st).size() - 2]);
  }
  return res;
}
int32_t main() {
  setIO();
  long long n;
  cin >> n;
  vector<long long> v;
  read(v, n);
  long long ans = solve(v);
  reverse(v.begin(), v.end());
  cout << max(ans, solve(v)) << '\n';
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
void setIO(const string &name = "") {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  if (name.length()) {
    freopen((name + ".in").c_str(), "r", stdin);
    freopen((name + ".out").c_str(), "w", stdout);
  }
}
template <typename T>
void read(vector<T> &a) {
  for (auto &x : a) cin >> x;
}
template <typename T>
void read(vector<T> &a, long long n) {
  a.resize(n);
  for (auto &x : a) cin >> x;
}
template <typename T>
void print(vector<T> &a) {
  for (const auto &x : a) cout << x << ' ';
  cout << '\n';
}
void MOD(long long &x) {
  x %= 1000000007;
  if (x < 0) x += 1000000007;
}
long long gcd(long long a, long long b) { return !a ? b : gcd(b % a, a); }
long long solve(const vector<long long> &v) {
  long long res = v[0] xor v[1];
  long long n = (long long)(v).size();
  vector<long long> st;
  for (auto(i) = 0; i < n; i++) {
    while (!st.empty() and st.back() < v[i]) st.pop_back();
    st.emplace_back(v[i]);
    if ((long long)(st).size() >= 2)
      res = max(res, st[(long long)(st).size() - 1] xor
                         st[(long long)(st).size() - 2]);
  }
  return res;
}
int32_t main() {
  setIO();
  long long n;
  cin >> n;
  vector<long long> v;
  read(v, n);
  long long ans = solve(v);
  reverse(v.begin(), v.end());
  cout << max(ans, solve(v)) << '\n';
  return 0;
}

```