#include <bits/stdc++.h>
using namespace std;
const int INF = 0x20202020;
const int MOD = 1000000007;
inline void read(long long &x) {
  char ch;
  bool f = 0;
  long long a = 0;
  while (!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-')))
    ;
  if (ch != '-')
    a *= 10, a += ch - '0';
  else
    f = 1;
  while (((ch = getchar()) >= '0') && (ch <= '9')) a *= 10, a += ch - '0';
  if (f) a = -a;
  x = a;
}
const int DX[] = {1, 0, -1, 0}, DY[] = {0, 1, 0, -1};
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= MOD;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
  }
  return res;
}
long long powmod(long long a, long long b, long long MOD) {
  long long res = 1;
  a %= MOD;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void RunCode() {
  long long N;
  read(N);
  stack<long long> Stack;
  long long ans = 0;
  long long s;
  for (int i = 1; i <= N; ++i) {
    read(s);
    while (!Stack.empty() && Stack.top() < s) {
      ans = max(ans, s ^ Stack.top());
      Stack.pop();
    }
    if (!Stack.empty()) {
      ans = max(ans, s ^ Stack.top());
    }
    Stack.push(s);
  }
  if (!Stack.empty()) {
    long long prev = Stack.top();
    Stack.pop();
    while (!Stack.empty()) {
      ans = max(ans, prev ^ Stack.top());
      prev = Stack.top();
      Stack.pop();
    }
  }
  cout << ans << endl;
}
int main() {
  long long T = 1;
  while (T--) RunCode();
  return 0;
}

### Prompt

Your task is to create 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;
const int INF = 0x20202020;
const int MOD = 1000000007;
inline void read(long long &x) {
  char ch;
  bool f = 0;
  long long a = 0;
  while (!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-')))
    ;
  if (ch != '-')
    a *= 10, a += ch - '0';
  else
    f = 1;
  while (((ch = getchar()) >= '0') && (ch <= '9')) a *= 10, a += ch - '0';
  if (f) a = -a;
  x = a;
}
const int DX[] = {1, 0, -1, 0}, DY[] = {0, 1, 0, -1};
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= MOD;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
  }
  return res;
}
long long powmod(long long a, long long b, long long MOD) {
  long long res = 1;
  a %= MOD;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void RunCode() {
  long long N;
  read(N);
  stack<long long> Stack;
  long long ans = 0;
  long long s;
  for (int i = 1; i <= N; ++i) {
    read(s);
    while (!Stack.empty() && Stack.top() < s) {
      ans = max(ans, s ^ Stack.top());
      Stack.pop();
    }
    if (!Stack.empty()) {
      ans = max(ans, s ^ Stack.top());
    }
    Stack.push(s);
  }
  if (!Stack.empty()) {
    long long prev = Stack.top();
    Stack.pop();
    while (!Stack.empty()) {
      ans = max(ans, prev ^ Stack.top());
      prev = Stack.top();
      Stack.pop();
    }
  }
  cout << ans << endl;
}
int main() {
  long long T = 1;
  while (T--) RunCode();
  return 0;
}

```