#include <bits/stdc++.h>
using namespace std;
long long bin_pow(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2 == 0) {
    long long t = bin_pow(a, b / 2);
    return t * t % 1000000007;
  } else
    return a * bin_pow(a, b - 1) % 1000000007;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1, n, m, k = 0, x = 0, y = 0, z = 0, sum = 0, l = 0, r = 0,
            ans = 0, mn = LLONG_MAX, mx = LLONG_MIN;
  cin >> n;
  vector<long long> a(n), b(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  stack<long long> s;
  for (int i = 0; i < n; i++) {
    while (!s.empty() && s.top() < a[i]) s.pop();
    if (s.size() >= 1) mx = max(mx, a[i] ^ s.top());
    s.push(a[i]);
  }
  reverse(a.begin(), a.end());
  stack<long long> s1;
  for (int i = 0; i < n; i++) {
    while (!s1.empty() && s1.top() < a[i]) s1.pop();
    if (s1.size() >= 1) mx = max(mx, a[i] ^ s1.top());
    s1.push(a[i]);
  }
  cout << mx;
  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 bin_pow(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2 == 0) {
    long long t = bin_pow(a, b / 2);
    return t * t % 1000000007;
  } else
    return a * bin_pow(a, b - 1) % 1000000007;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1, n, m, k = 0, x = 0, y = 0, z = 0, sum = 0, l = 0, r = 0,
            ans = 0, mn = LLONG_MAX, mx = LLONG_MIN;
  cin >> n;
  vector<long long> a(n), b(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  stack<long long> s;
  for (int i = 0; i < n; i++) {
    while (!s.empty() && s.top() < a[i]) s.pop();
    if (s.size() >= 1) mx = max(mx, a[i] ^ s.top());
    s.push(a[i]);
  }
  reverse(a.begin(), a.end());
  stack<long long> s1;
  for (int i = 0; i < n; i++) {
    while (!s1.empty() && s1.top() < a[i]) s1.pop();
    if (s1.size() >= 1) mx = max(mx, a[i] ^ s1.top());
    s1.push(a[i]);
  }
  cout << mx;
  return 0;
}

```