#include <bits/stdc++.h>
using namespace std;
int ToInt(string& s, int len) {
  int r = 0;
  for (int i = 0, max_ = (len); i < max_; ++i) {
    r *= 10;
    r += s[i] - '0';
  }
  return r;
}
int GCD(int a, int b) { return b != 0 ? GCD(b, a % b) : a; }
int LCM(int a, int b) { return a * (b / GCD(a, b)); }
long long Pow(long long n, long long e) {
  if (e == 0) return 1;
  if (e == 1)
    return n;
  else if (e & 1) {
    long long t = Pow(n, e / 2);
    return n * t * t;
  } else {
    long long t = Pow(n, e / 2);
    return t * t;
  }
}
int main() {
  vector<int> v;
  int N, res = 0;
  scanf("%d", &N);
  for (int i = 0, max_ = (N); i < max_; ++i) {
    int n;
    scanf("%d", &n);
    while ((int)(v).size() > 0) {
      res = max(res, v[(int)(v).size() - 1] ^ n);
      if (v[(int)(v).size() - 1] > n) break;
      v.pop_back();
    }
    v.push_back(n);
  }
  printf("%d\n", res);
  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;
int ToInt(string& s, int len) {
  int r = 0;
  for (int i = 0, max_ = (len); i < max_; ++i) {
    r *= 10;
    r += s[i] - '0';
  }
  return r;
}
int GCD(int a, int b) { return b != 0 ? GCD(b, a % b) : a; }
int LCM(int a, int b) { return a * (b / GCD(a, b)); }
long long Pow(long long n, long long e) {
  if (e == 0) return 1;
  if (e == 1)
    return n;
  else if (e & 1) {
    long long t = Pow(n, e / 2);
    return n * t * t;
  } else {
    long long t = Pow(n, e / 2);
    return t * t;
  }
}
int main() {
  vector<int> v;
  int N, res = 0;
  scanf("%d", &N);
  for (int i = 0, max_ = (N); i < max_; ++i) {
    int n;
    scanf("%d", &n);
    while ((int)(v).size() > 0) {
      res = max(res, v[(int)(v).size() - 1] ^ n);
      if (v[(int)(v).size() - 1] > n) break;
      v.pop_back();
    }
    v.push_back(n);
  }
  printf("%d\n", res);
  return 0;
}

```