#include <bits/stdc++.h>
using namespace std;
int N;
int s[100005], lef[100005], rgt[100005];
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", &s[i]);
  stack<int> stk;
  lef[0] = -1;
  stk.push(s[0]);
  for (int i = 1; i < N; i++) {
    while (!stk.empty() && stk.top() < s[i]) stk.pop();
    if (stk.empty())
      lef[i] = -1;
    else
      lef[i] = stk.top();
    stk.push(s[i]);
  }
  while (!stk.empty()) stk.pop();
  rgt[N - 1] = -1;
  stk.push(s[N - 1]);
  for (int i = N - 2; i >= 0; i--) {
    while (!stk.empty() && stk.top() < s[i]) stk.pop();
    if (stk.empty())
      rgt[i] = -1;
    else
      rgt[i] = stk.top();
    stk.push(s[i]);
  }
  while (!stk.empty()) stk.pop();
  int ans = 0;
  for (int i = 0; i < N; i++) {
    if (lef[i] != -1) ans = max(ans, s[i] ^ lef[i]);
    if (rgt[i] != -1) ans = max(ans, s[i] ^ rgt[i]);
  }
  printf("%d\n", ans);
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
int N;
int s[100005], lef[100005], rgt[100005];
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", &s[i]);
  stack<int> stk;
  lef[0] = -1;
  stk.push(s[0]);
  for (int i = 1; i < N; i++) {
    while (!stk.empty() && stk.top() < s[i]) stk.pop();
    if (stk.empty())
      lef[i] = -1;
    else
      lef[i] = stk.top();
    stk.push(s[i]);
  }
  while (!stk.empty()) stk.pop();
  rgt[N - 1] = -1;
  stk.push(s[N - 1]);
  for (int i = N - 2; i >= 0; i--) {
    while (!stk.empty() && stk.top() < s[i]) stk.pop();
    if (stk.empty())
      rgt[i] = -1;
    else
      rgt[i] = stk.top();
    stk.push(s[i]);
  }
  while (!stk.empty()) stk.pop();
  int ans = 0;
  for (int i = 0; i < N; i++) {
    if (lef[i] != -1) ans = max(ans, s[i] ^ lef[i]);
    if (rgt[i] != -1) ans = max(ans, s[i] ^ rgt[i]);
  }
  printf("%d\n", ans);
}

```