#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  long long n, A[100001], MaxR[100001], MaxL[100001], i, x, r, l, best;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> A[i];
  }
  MaxL[0] = -1;
  MaxR[n - 1] = n;
  for (i = 1; i < n; i++) {
    x = i - 1;
    while (x != -1 && A[x] <= A[i]) {
      x = MaxL[x];
    }
    MaxL[i] = x;
  }
  for (i = n - 2; i >= 0; i--) {
    x = i + 1;
    while (x != n && A[x] <= A[i]) {
      x = MaxR[x];
    }
    MaxR[i] = x;
  }
  best = 0;
  for (i = 0; i < n; i++) {
    r = MaxR[i];
    l = MaxL[i];
    if (r == n && l == -1) {
      continue;
    }
    if (r == n) {
      best = max(best, A[i] ^ A[l]);
      continue;
    }
    if (l == -1) {
      best = max(best, A[i] ^ A[r]);
      continue;
    }
    best = max(best, A[i] ^ A[l]);
    best = max(best, A[i] ^ A[r]);
    best = max(best, A[l] ^ A[r]);
  }
  cout << best;
}

### Prompt

Please create a solution in cpp 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;
int main() {
  ios::sync_with_stdio(false);
  long long n, A[100001], MaxR[100001], MaxL[100001], i, x, r, l, best;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> A[i];
  }
  MaxL[0] = -1;
  MaxR[n - 1] = n;
  for (i = 1; i < n; i++) {
    x = i - 1;
    while (x != -1 && A[x] <= A[i]) {
      x = MaxL[x];
    }
    MaxL[i] = x;
  }
  for (i = n - 2; i >= 0; i--) {
    x = i + 1;
    while (x != n && A[x] <= A[i]) {
      x = MaxR[x];
    }
    MaxR[i] = x;
  }
  best = 0;
  for (i = 0; i < n; i++) {
    r = MaxR[i];
    l = MaxL[i];
    if (r == n && l == -1) {
      continue;
    }
    if (r == n) {
      best = max(best, A[i] ^ A[l]);
      continue;
    }
    if (l == -1) {
      best = max(best, A[i] ^ A[r]);
      continue;
    }
    best = max(best, A[i] ^ A[l]);
    best = max(best, A[i] ^ A[r]);
    best = max(best, A[l] ^ A[r]);
  }
  cout << best;
}

```