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}.
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;
const long long infL = 1e18;
int setbit(int n, int pos) { return n = n | (1 << pos); }
int resetbit(int n, int pos) { return n = n & ~(1 << pos); }
bool checkbit(int n, int pos) { return (bool)(n & (1 << pos)); }
template <typename T>
T mul(T x, T y) {
  x %= mod;
  y %= mod;
  return (x * y) % mod;
}
template <typename T>
T add(T x, T y) {
  x %= mod;
  y %= mod;
  return (x + y) % mod;
}
template <typename T>
T modPow(T x, T k, T mod) {
  if (k == 0) return 1;
  if (k & 1) {
    return ((x % mod) * modPow(x, k - 1, mod) % mod) % mod;
  } else {
    T ret = modPow(x, k / 2, mod);
    ret %= mod;
    return (ret * ret) % mod;
  }
}
template <typename T>
T ext_gcd(T num1, T num2, T &X, T &Y) {
  T r2, r1, q2, q1, x2, x1, y2, y1, x, y, r;
  x2 = 1;
  y2 = 0;
  x1 = 0;
  y1 = 1;
  for (r2 = num1, r1 = num2; r1 != 0;
       y2 = y1, y1 = y, x2 = x1, x1 = x, r2 = r1, r1 = r) {
    q1 = r2 / r1;
    x = x2 - q1 * x1;
    y = y2 - q1 * y1;
    r = r2 % r1;
  }
  X = x2;
  Y = y2;
  return r2;
}
template <typename T>
T modInv(T x, T m) {
  return modPow(x, m - 2, m);
}
template <typename T>
void print(const T &v) {
  cerr << v << ' ';
}
template <typename T1, typename... T2>
void print(const T1 &first, const T2 &...rest) {
  print(first);
  print(rest...);
}
template <typename F, typename S>
ostream &operator<<(ostream &os, const pair<F, S> &p) {
  return os << "[ " << p.first << ", " << p.second << " ]";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "[ ";
  for (int i = 0; i < v.size(); i++) os << v[i] << ' ';
  return os << " ]";
}
int modAdd(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
const int M = (int)3e5 + 7;
const int mx = (int)5000 + 123;
const int nx = 500 + 2;
int a[nx + 2];
vector<long long> v;
void solve() {
  long long n, k, x, ans = 0LL;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x;
    while (v.size() and v.back() <= x) {
      ans = max(ans, v.back() ^ x);
      v.pop_back();
    }
    if (v.size() and v.back() > x) {
      ans = max(v.back() ^ x, ans);
    }
    v.push_back(x);
  }
  cout << ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1, c = 0;
  while (t--) {
    solve();
  }
}
