#include <bits/stdc++.h>
const long long int mod = 1e9 + 7;
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool IsPowerOfTwo(int x) { return (x != 0) && ((x & (x - 1)) == 0); }
long long int powm(int a, int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * 1LL * a) % mod;
    b >>= 1;
    a = (a * 1LL * a) % mod;
  }
  return res;
}
long long int query(long long int i, long long int j) {
  cout << "? " << i << ' ' << j << endl;
  int x;
  cin >> x;
  return x;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  int a[n];
  for (int(i) = 0; (i) < (n); ++(i)) cin >> a[i];
  stack<int> st;
  int b[n];
  for (int i = n - 1; i >= 0; i--) {
    while (!st.empty() && st.top() < a[i]) {
      st.pop();
    }
    if (st.empty())
      b[i] = -1;
    else
      b[i] = st.top();
    st.push(a[i]);
  }
  int ans = 0;
  for (int(i) = 0; (i) < (n); ++(i)) {
    if (b[i] == -1) continue;
    ans = max(ans, a[i] ^ b[i]);
  }
  memset(b, 0, sizeof(b));
  for (int i = 0; i < n; i++) {
    while (!st.empty() && st.top() < a[i]) {
      st.pop();
    }
    if (st.empty())
      b[i] = -1;
    else
      b[i] = st.top();
    st.push(a[i]);
  }
  for (int(i) = 0; (i) < (n); ++(i)) {
    if (b[i] == -1) continue;
    ans = max(ans, a[i] ^ b[i]);
  }
  cout << 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>
const long long int mod = 1e9 + 7;
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool IsPowerOfTwo(int x) { return (x != 0) && ((x & (x - 1)) == 0); }
long long int powm(int a, int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * 1LL * a) % mod;
    b >>= 1;
    a = (a * 1LL * a) % mod;
  }
  return res;
}
long long int query(long long int i, long long int j) {
  cout << "? " << i << ' ' << j << endl;
  int x;
  cin >> x;
  return x;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  int a[n];
  for (int(i) = 0; (i) < (n); ++(i)) cin >> a[i];
  stack<int> st;
  int b[n];
  for (int i = n - 1; i >= 0; i--) {
    while (!st.empty() && st.top() < a[i]) {
      st.pop();
    }
    if (st.empty())
      b[i] = -1;
    else
      b[i] = st.top();
    st.push(a[i]);
  }
  int ans = 0;
  for (int(i) = 0; (i) < (n); ++(i)) {
    if (b[i] == -1) continue;
    ans = max(ans, a[i] ^ b[i]);
  }
  memset(b, 0, sizeof(b));
  for (int i = 0; i < n; i++) {
    while (!st.empty() && st.top() < a[i]) {
      st.pop();
    }
    if (st.empty())
      b[i] = -1;
    else
      b[i] = st.top();
    st.push(a[i]);
  }
  for (int(i) = 0; (i) < (n); ++(i)) {
    if (b[i] == -1) continue;
    ans = max(ans, a[i] ^ b[i]);
  }
  cout << ans;
  return 0;
}

```