#include <bits/stdc++.h>
using namespace std;
int n, a[100005], ls[100005], t[200005];
int lowbit(int x) { return x & (-x); }
void insert1(int now, int val) {
  while (now) {
    t[now] = max(t[now], val);
    now -= lowbit(now);
  }
}
void insert2(int now, int val) {
  while (now) {
    t[now] = min(t[now], val);
    now -= lowbit(now);
  }
}
int ask1(int now) {
  int re = 0;
  while (now <= n) {
    re = max(re, t[now]);
    now += lowbit(now);
  }
  return re;
}
int ask2(int now) {
  int re = n + 1;
  while (now <= n) {
    re = min(re, t[now]);
    now += lowbit(now);
  }
  return re;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), ls[i] = a[i];
  if (n == 1) {
    printf("%d\n", 0);
    return 0;
  }
  sort(ls + 1, ls + n + 1);
  ls[0] = unique(ls + 1, ls + n + 1) - ls - 1;
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(ls + 1, ls + ls[0] + 1, a[i]) - ls;
  int ans = 0;
  insert1(a[1], 1);
  for (int i = 2; i <= n; i++) {
    if (ask1(a[i] + 1) > 0) ans = max(ans, ls[a[i]] ^ ls[a[ask1(a[i] + 1)]]);
    insert1(a[i], i);
  }
  for (int i = 1; i <= n; i++) t[i] = n + 1;
  insert2(a[n], n);
  for (int i = n - 1; i >= 1; i--) {
    if (ask2(a[i] + 1) < n + 1)
      ans = max(ans, ls[a[i]] ^ ls[a[ask2(a[i] + 1)]]);
    insert2(a[i], i);
  }
  printf("%d\n", ans);
  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 n, a[100005], ls[100005], t[200005];
int lowbit(int x) { return x & (-x); }
void insert1(int now, int val) {
  while (now) {
    t[now] = max(t[now], val);
    now -= lowbit(now);
  }
}
void insert2(int now, int val) {
  while (now) {
    t[now] = min(t[now], val);
    now -= lowbit(now);
  }
}
int ask1(int now) {
  int re = 0;
  while (now <= n) {
    re = max(re, t[now]);
    now += lowbit(now);
  }
  return re;
}
int ask2(int now) {
  int re = n + 1;
  while (now <= n) {
    re = min(re, t[now]);
    now += lowbit(now);
  }
  return re;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), ls[i] = a[i];
  if (n == 1) {
    printf("%d\n", 0);
    return 0;
  }
  sort(ls + 1, ls + n + 1);
  ls[0] = unique(ls + 1, ls + n + 1) - ls - 1;
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(ls + 1, ls + ls[0] + 1, a[i]) - ls;
  int ans = 0;
  insert1(a[1], 1);
  for (int i = 2; i <= n; i++) {
    if (ask1(a[i] + 1) > 0) ans = max(ans, ls[a[i]] ^ ls[a[ask1(a[i] + 1)]]);
    insert1(a[i], i);
  }
  for (int i = 1; i <= n; i++) t[i] = n + 1;
  insert2(a[n], n);
  for (int i = n - 1; i >= 1; i--) {
    if (ask2(a[i] + 1) < n + 1)
      ans = max(ans, ls[a[i]] ^ ls[a[ask2(a[i] + 1)]]);
    insert2(a[i], i);
  }
  printf("%d\n", ans);
  return 0;
}

```