#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e5 + 5;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long n, a[maxn];
stack<long long> s;
long long ans;
inline long long solve() {
  long long ret = 0;
  for (long long i = 1; i <= n; i++) {
    while (s.size() && s.top() < a[i]) s.pop();
    if (s.size()) ret = max(ret, s.top() ^ a[i]);
    s.push(a[i]);
  }
  return ret;
}
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  ans = solve();
  reverse(a + 1, a + n + 1);
  ans = max(ans, solve());
  cout << ans;
  return 0;
}

### Prompt

Generate 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;
const long long maxn = 1e5 + 5;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long n, a[maxn];
stack<long long> s;
long long ans;
inline long long solve() {
  long long ret = 0;
  for (long long i = 1; i <= n; i++) {
    while (s.size() && s.top() < a[i]) s.pop();
    if (s.size()) ret = max(ret, s.top() ^ a[i]);
    s.push(a[i]);
  }
  return ret;
}
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  ans = solve();
  reverse(a + 1, a + n + 1);
  ans = max(ans, solve());
  cout << ans;
  return 0;
}

```