#include <bits/stdc++.h>
using namespace std;
const int N = 1e6;
int n, t[1000001];
struct data {
  int num, loc, vis;
} b[1000001];
int a[1000001];
int lowbit(int x) { return x & (-x); }
int Query(int x) {
  int sum = 0;
  while (x > 0) {
    sum = max(sum, t[x]);
    x -= lowbit(x);
  }
  return sum;
}
void Add(int x, int va) {
  while (x <= N) {
    t[x] = max(t[x], va);
    x += lowbit(x);
  }
}
bool cmp(data p, data q) { return p.num < q.num; }
bool cmp2(data p, data q) { return p.vis < q.vis; }
int work() {
  memset(t, 0, sizeof(t));
  for (int i = 1; i <= n; i++) {
    b[i].num = a[i];
    b[i].vis = i;
  }
  sort(b + 1, b + n + 1, cmp);
  for (int i = 1; i <= n; i++) b[i].loc = n + 1 - i;
  sort(b + 1, b + n + 1, cmp2);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int nowans = Query(b[i].loc);
    Add(b[i].loc, i);
    if (nowans == 0) continue;
    ans = max(ans, b[i].num ^ b[nowans].num);
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int ans = work();
  for (int i = 1; i <= n / 2; i++) swap(a[i], a[n + 1 - i]);
  ans = max(ans, work());
  printf("%d\n", ans);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
const int N = 1e6;
int n, t[1000001];
struct data {
  int num, loc, vis;
} b[1000001];
int a[1000001];
int lowbit(int x) { return x & (-x); }
int Query(int x) {
  int sum = 0;
  while (x > 0) {
    sum = max(sum, t[x]);
    x -= lowbit(x);
  }
  return sum;
}
void Add(int x, int va) {
  while (x <= N) {
    t[x] = max(t[x], va);
    x += lowbit(x);
  }
}
bool cmp(data p, data q) { return p.num < q.num; }
bool cmp2(data p, data q) { return p.vis < q.vis; }
int work() {
  memset(t, 0, sizeof(t));
  for (int i = 1; i <= n; i++) {
    b[i].num = a[i];
    b[i].vis = i;
  }
  sort(b + 1, b + n + 1, cmp);
  for (int i = 1; i <= n; i++) b[i].loc = n + 1 - i;
  sort(b + 1, b + n + 1, cmp2);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int nowans = Query(b[i].loc);
    Add(b[i].loc, i);
    if (nowans == 0) continue;
    ans = max(ans, b[i].num ^ b[nowans].num);
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int ans = work();
  for (int i = 1; i <= n / 2; i++) swap(a[i], a[n + 1 - i]);
  ans = max(ans, work());
  printf("%d\n", ans);
  return 0;
}

```