#include <bits/stdc++.h>
using namespace std;
int a[100005];
int n;
int ml;
void work(int st, int nd) {
  if (nd - st + 1 < 2) {
    return;
  }
  int m = (st + nd) / 2;
  int sec;
  if (m > st) {
    sec = a[m - 1];
    for (int i = m - 1; i >= st; i--) {
      if (a[i] > sec) {
        sec = a[i];
      }
      if ((sec ^ a[m]) > ml) {
        ml = (sec ^ a[m]);
      }
      if (sec > a[m]) {
        break;
      }
    }
    work(st, m - 1);
  }
  if (m < nd) {
    sec = a[m + 1];
    for (int i = m + 1; i <= nd; i++) {
      if (a[i] > sec) {
        sec = a[i];
      }
      if ((sec ^ a[m]) > ml) {
        ml = (sec ^ a[m]);
      }
      if (sec > a[m]) {
        break;
      }
    }
    work(m + 1, nd);
  }
  int l, r;
  int la, ra;
  l = m - 1;
  r = m + 1;
  while (l >= st && a[l] < a[m]) {
    l--;
  }
  if (l >= st) {
    la = a[l];
  }
  while (r <= nd && a[r] < a[m]) {
    r++;
  }
  if (r <= nd) {
    ra = a[r];
  }
  while (l >= st && r <= nd) {
    if ((a[l] ^ a[r]) > ml) {
      ml = (a[l] ^ a[r]);
    }
    if (a[l] < a[r]) {
      while (l >= st && a[l] <= la) {
        l--;
      }
      if (l >= st) {
        la = a[l];
      }
    } else {
      while (r <= nd && a[r] <= ra) {
        r++;
      }
      if (r <= nd) {
        ra = a[r];
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  ml = -1;
  work(1, n);
  printf("%d\n", ml);
  return 0;
}

### 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 a[100005];
int n;
int ml;
void work(int st, int nd) {
  if (nd - st + 1 < 2) {
    return;
  }
  int m = (st + nd) / 2;
  int sec;
  if (m > st) {
    sec = a[m - 1];
    for (int i = m - 1; i >= st; i--) {
      if (a[i] > sec) {
        sec = a[i];
      }
      if ((sec ^ a[m]) > ml) {
        ml = (sec ^ a[m]);
      }
      if (sec > a[m]) {
        break;
      }
    }
    work(st, m - 1);
  }
  if (m < nd) {
    sec = a[m + 1];
    for (int i = m + 1; i <= nd; i++) {
      if (a[i] > sec) {
        sec = a[i];
      }
      if ((sec ^ a[m]) > ml) {
        ml = (sec ^ a[m]);
      }
      if (sec > a[m]) {
        break;
      }
    }
    work(m + 1, nd);
  }
  int l, r;
  int la, ra;
  l = m - 1;
  r = m + 1;
  while (l >= st && a[l] < a[m]) {
    l--;
  }
  if (l >= st) {
    la = a[l];
  }
  while (r <= nd && a[r] < a[m]) {
    r++;
  }
  if (r <= nd) {
    ra = a[r];
  }
  while (l >= st && r <= nd) {
    if ((a[l] ^ a[r]) > ml) {
      ml = (a[l] ^ a[r]);
    }
    if (a[l] < a[r]) {
      while (l >= st && a[l] <= la) {
        l--;
      }
      if (l >= st) {
        la = a[l];
      }
    } else {
      while (r <= nd && a[r] <= ra) {
        r++;
      }
      if (r <= nd) {
        ra = a[r];
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  ml = -1;
  work(1, n);
  printf("%d\n", ml);
  return 0;
}

```