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>
using namespace std;
const long long mod = 998244353;
long long np2(long long i) {
  if ((i & (i - 1)) == 0) {
    return i;
  }
  while ((i & (i - 1))) {
    i = (i & (i - 1));
  }
  return i << 1;
}
vector<long long> segtree;
vector<long long> a;
long long n;
long long ans = 0;
void createTree(long long lo, long long hi, long long pos) {
  if (lo == hi) {
    segtree[pos] = lo;
    return;
  }
  long long mid = (lo + hi) / 2;
  createTree(lo, mid, 2 * pos + 1);
  createTree(mid + 1, hi, 2 * pos + 2);
  if (a[segtree[2 * pos + 1]] > a[segtree[2 * pos + 2]]) {
    segtree[pos] = segtree[2 * pos + 1];
  } else {
    segtree[pos] = segtree[2 * pos + 2];
  }
  return;
}
long long rangeQuery(long long qlo, long long qhi, long long lo, long long hi,
                     long long pos) {
  if (lo > qhi || hi < qlo) {
    return n;
  }
  if (lo >= qlo && hi <= qhi) {
    return segtree[pos];
  }
  long long mid = (hi + lo) / 2;
  long long left = rangeQuery(qlo, qhi, lo, mid, 2 * pos + 1);
  long long right = rangeQuery(qlo, qhi, mid + 1, hi, 2 * pos + 2);
  if (a[right] > a[left]) {
    left = right;
  }
  return left;
}
long long func(long long lo, long long hi, long long mx) {
  if (hi <= lo) {
    return 0;
  }
  long long left = mx > lo ? rangeQuery(lo, mx - 1, 0, n - 1, 0) : n;
  long long right = mx < hi ? rangeQuery(mx + 1, hi, 0, n - 1, 0) : n;
  long long ans = max(a[left], a[right]);
  ans = (ans ^ a[mx]);
  ans = max(ans, func(lo, mx - 1, left));
  ans = max(ans, func(mx + 1, hi, right));
  return ans;
}
void fill(long long mx, long long lo, long long hi) {
  if (mx == n || hi <= lo) {
    return;
  }
  long long left = mx > lo ? rangeQuery(lo, mx - 1, 0, n - 1, 0) : n;
  long long right = mx < hi ? rangeQuery(mx + 1, hi, 0, n - 1, 0) : n;
  fill(left, lo, mx - 1);
  fill(right, mx + 1, hi);
  while (left != n) {
    ans = max(ans, a[mx] ^ a[left]);
    lo = left + 1;
    left = mx > lo ? rangeQuery(lo, mx - 1, 0, n - 1, 0) : n;
  }
  while (right != n) {
    ans = max(ans, a[mx] ^ a[right]);
    hi = right - 1;
    right = mx < hi ? rangeQuery(mx + 1, hi, 0, n - 1, 0) : n;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  segtree.resize(2 * np2(n) - 1, n);
  a.resize(n + 1);
  for (long long i = (long long)0; i != (long long)n; i = i + 1) {
    cin >> a[i];
  }
  a[n] = 0;
  createTree(0, n - 1, 0);
  fill(segtree[0], 0, n - 1);
  cout << ans << '\n';
}
