#include <bits/stdc++.h>
using namespace std;
const long long mx = 100001;
const long long mod = 1000000007;
const long long inf = 0x3f3f3f3f;
void solve();
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
    if (i != t) printf("\n");
  }
  return 0;
}
long long ar[mx], D[mx];
long long ma, n;
void sol() {
  int i, j, k, m;
  stack<int> s1;
  for (int i = 0; i <= n - 1; i++) {
    if (s1.empty())
      s1.push(i);
    else {
      while (!s1.empty() && ar[s1.top()] <= ar[i]) {
        D[s1.top()] = i;
        s1.pop();
      }
      s1.push(i);
    }
  }
  while (!s1.empty()) {
    D[s1.top()] = -1;
    s1.pop();
  }
  for (int i = 0; i <= n - 1; i++) {
    long long x = D[i] == -1 ? -1 : ar[D[i]];
    ma = max(ma, ar[i] ^ x);
  }
}
void solve() {
  int i, j, k, m;
  stack<int> s1;
  scanf("%lld", &n);
  for (int i = 0; i <= n - 1; i++) scanf("%lld", &ar[i]);
  sol();
  reverse(ar, ar + n);
  sol();
  cout << ma;
}

### 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 mx = 100001;
const long long mod = 1000000007;
const long long inf = 0x3f3f3f3f;
void solve();
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
    if (i != t) printf("\n");
  }
  return 0;
}
long long ar[mx], D[mx];
long long ma, n;
void sol() {
  int i, j, k, m;
  stack<int> s1;
  for (int i = 0; i <= n - 1; i++) {
    if (s1.empty())
      s1.push(i);
    else {
      while (!s1.empty() && ar[s1.top()] <= ar[i]) {
        D[s1.top()] = i;
        s1.pop();
      }
      s1.push(i);
    }
  }
  while (!s1.empty()) {
    D[s1.top()] = -1;
    s1.pop();
  }
  for (int i = 0; i <= n - 1; i++) {
    long long x = D[i] == -1 ? -1 : ar[D[i]];
    ma = max(ma, ar[i] ^ x);
  }
}
void solve() {
  int i, j, k, m;
  stack<int> s1;
  scanf("%lld", &n);
  for (int i = 0; i <= n - 1; i++) scanf("%lld", &ar[i]);
  sol();
  reverse(ar, ar + n);
  sol();
  cout << ma;
}

```