#include <bits/stdc++.h>
using namespace std;
inline int max(int a, int b) { return a > b ? a : b; }
inline int min(int a, int b) { return a < b ? a : b; }
inline int gcd(int a, int b) {
  if (a % b)
    return gcd(b, a % b);
  else
    return b;
}
inline int lcm(int a, int b) { return (a * (b / gcd(a, b))); }
inline int And(int mask, int bit) { return mask & (1 << bit); }
inline int Or(int mask, int bit) { return mask | (1 << bit); }
inline int Xor(int mask, int bit) { return mask & (~(1 << bit)); }
int n, ans, x, y;
int a[100111], l[100111], r[100111];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  l[1] = 0;
  r[n] = n + 1;
  for (int i = 2; i <= n; i++) {
    l[i] = i - 1;
    while (a[l[i]] < a[i]) {
      l[i] = l[l[i]];
      if (l[i] == 0) break;
    }
  }
  for (int i = n - 1; i >= 1; i--) {
    r[i] = i + 1;
    while (a[r[i]] < a[i]) {
      r[i] = r[r[i]];
      if (r[i] == n + 1) break;
    }
  }
  ans = 0;
  for (int i = 1; i <= n; i++) {
    x = l[i];
    y = r[i];
    if (x != i && x != 0) ans = max(ans, a[i] ^ a[x]);
    if (y != i && y != n + 1) ans = max(ans, a[i] ^ a[y]);
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate 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;
inline int max(int a, int b) { return a > b ? a : b; }
inline int min(int a, int b) { return a < b ? a : b; }
inline int gcd(int a, int b) {
  if (a % b)
    return gcd(b, a % b);
  else
    return b;
}
inline int lcm(int a, int b) { return (a * (b / gcd(a, b))); }
inline int And(int mask, int bit) { return mask & (1 << bit); }
inline int Or(int mask, int bit) { return mask | (1 << bit); }
inline int Xor(int mask, int bit) { return mask & (~(1 << bit)); }
int n, ans, x, y;
int a[100111], l[100111], r[100111];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  l[1] = 0;
  r[n] = n + 1;
  for (int i = 2; i <= n; i++) {
    l[i] = i - 1;
    while (a[l[i]] < a[i]) {
      l[i] = l[l[i]];
      if (l[i] == 0) break;
    }
  }
  for (int i = n - 1; i >= 1; i--) {
    r[i] = i + 1;
    while (a[r[i]] < a[i]) {
      r[i] = r[r[i]];
      if (r[i] == n + 1) break;
    }
  }
  ans = 0;
  for (int i = 1; i <= n; i++) {
    x = l[i];
    y = r[i];
    if (x != i && x != 0) ans = max(ans, a[i] ^ a[x]);
    if (y != i && y != n + 1) ans = max(ans, a[i] ^ a[y]);
  }
  printf("%d\n", ans);
  return 0;
}

```