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 int MAXN = 1 << 17;
int n, s[MAXN], pl[MAXN], pr[MAXN], ans = 0;
queue<int> q;
inline int maxId(const int& i, const int& j) { return (s[i] > s[j]) ? i : j; }
struct RQTree {
  int a[2 * MAXN];
  RQTree() { memset(a, -1, sizeof a); }
  int update(int x, int y, int i = 1, int l = 0, int r = MAXN - 1) {
    if (y < l || r < x || l == r) return a[i];
    int m = (l + r) / 2;
    return a[i] = maxId(update(x, y, 2 * i, l, m),
                        update(x, y, 2 * i + 1, m + 1, r));
  }
  int query(int x, int y, int i = 1, int l = 0, int r = MAXN - 1) {
    if (y < l || r < x) return -1;
    if (x <= l && r <= y) return a[i];
    int m = (l + r) / 2;
    return maxId(query(x, y, 2 * i, l, m), query(x, y, 2 * i + 1, m + 1, r));
  }
} tree;
void search() {
  while (!q.empty()) {
    int id = q.front();
    q.pop();
    if (pl[id] != 0) ans = max(ans, s[id] ^ s[pl[id]]);
    if (pr[id] != n + 1) ans = max(ans, s[id] ^ s[pr[id]]);
    if (id - pl[id] > 1) {
      int c = tree.query(pl[id] + 1, id - 1);
      pl[c] = pl[id], pr[c] = id, q.push(c);
    }
    if (pr[id] - id > 1) {
      int c = tree.query(id + 1, pr[id] - 1);
      pl[c] = id, pr[c] = pr[id], q.push(c);
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  memset(s, -1, sizeof s);
  s[0] = s[n + 1] = 0x3f3f3f3f;
  for (int i = 1; i <= n; ++i) cin >> s[i];
  for (int i = 0; i < MAXN; ++i) tree.a[MAXN + i] = i;
  tree.update(0, MAXN - 1);
  memset(pl, -1, sizeof pl);
  memset(pr, -1, sizeof pr);
  int bigId = tree.query(1, n);
  pl[bigId] = 0, pr[bigId] = n + 1;
  q.push(bigId);
  search();
  cout << ans << endl;
}
