#include <bits/stdc++.h>
using namespace std;
int n;
int a[100005], x[100005], r[100005];
int t[100005 * 10];
void init(int i, int l, int r) {
  t[i] = n + 1;
  int mid = (l + r) >> 1;
  if (l == r) return;
  init(i + i, l, mid);
  init(i + i + 1, mid + 1, r);
}
void insert(int i, int ln, int rn, int k, int v) {
  t[i] = min(t[i], v);
  if (ln == rn) return;
  int mid = (ln + rn) >> 1;
  if (k <= mid)
    insert(i + i, ln, mid, k, v);
  else
    insert(i + i + 1, mid + 1, rn, k, v);
}
int query(int i, int ln, int rn, int l, int r) {
  if (l <= ln && rn <= r) return t[i];
  int ret = n + 1, mid = (ln + rn) >> 1;
  if (l <= mid) ret = query(i + i, ln, mid, l, r);
  if (r > mid) ret = min(ret, query(i + i + 1, mid + 1, rn, l, r));
  return ret;
}
void gao() {
  for (int i = 1; i <= n; ++i) x[i] = a[i];
  sort(x + 1, x + n + 1);
  init(1, 1, n);
  for (int i = n; i >= 1; --i) {
    int k = lower_bound(x + 1, x + n + 1, a[i]) - x;
    r[i] = query(1, 1, n, k, n);
    insert(1, 1, n, k, i);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  gao();
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = i + 1; j <= n; j = r[j]) {
      ans = max(ans, a[i] ^ a[j]);
      if (a[i] < a[j]) break;
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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;
int n;
int a[100005], x[100005], r[100005];
int t[100005 * 10];
void init(int i, int l, int r) {
  t[i] = n + 1;
  int mid = (l + r) >> 1;
  if (l == r) return;
  init(i + i, l, mid);
  init(i + i + 1, mid + 1, r);
}
void insert(int i, int ln, int rn, int k, int v) {
  t[i] = min(t[i], v);
  if (ln == rn) return;
  int mid = (ln + rn) >> 1;
  if (k <= mid)
    insert(i + i, ln, mid, k, v);
  else
    insert(i + i + 1, mid + 1, rn, k, v);
}
int query(int i, int ln, int rn, int l, int r) {
  if (l <= ln && rn <= r) return t[i];
  int ret = n + 1, mid = (ln + rn) >> 1;
  if (l <= mid) ret = query(i + i, ln, mid, l, r);
  if (r > mid) ret = min(ret, query(i + i + 1, mid + 1, rn, l, r));
  return ret;
}
void gao() {
  for (int i = 1; i <= n; ++i) x[i] = a[i];
  sort(x + 1, x + n + 1);
  init(1, 1, n);
  for (int i = n; i >= 1; --i) {
    int k = lower_bound(x + 1, x + n + 1, a[i]) - x;
    r[i] = query(1, 1, n, k, n);
    insert(1, 1, n, k, i);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  gao();
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = i + 1; j <= n; j = r[j]) {
      ans = max(ans, a[i] ^ a[j]);
      if (a[i] < a[j]) break;
    }
  }
  printf("%d\n", ans);
  return 0;
}

```