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 maxint = -1u >> 1;
const double pi = 3.14159265358979323;
const double eps = 1e-8;
int n, val[100100], loc[100100], rk[100100];
int Max[100100][30], ans;
void initRMQ() {
  for (int i = 1; i <= n; i++) Max[i][0] = i;
  for (int k = 1; (1 << k) <= n; k++) {
    for (int i = 1; i + (1 << k) - 1 <= n; i++) {
      Max[i][k] = val[Max[i][k - 1]] > val[Max[i + (1 << (k - 1))][k - 1]]
                      ? Max[i][k - 1]
                      : Max[i + (1 << (k - 1))][k - 1];
    }
  }
}
int MaxRMQ(int a, int b) {
  if (a > b) return -1;
  int k = int(log(b - a + 1.0) / log(2.0));
  return val[Max[a][k]] > val[Max[b - (1 << k) + 1][k]]
             ? Max[a][k]
             : Max[b - (1 << k) + 1][k];
}
void solve(int l, int r) {
  if (l >= r) return;
  int bl = MaxRMQ(l, r);
  int lm = MaxRMQ(l, bl - 1);
  if (lm == -1) lm = bl;
  int rm = MaxRMQ(bl + 1, r);
  if (rm == -1) rm = bl;
  int mx = 0;
  for (int i = bl - 1; i >= lm; i--) {
    mx = max(mx, val[i]);
    ans = max(ans, val[bl] ^ mx);
  }
  mx = 0;
  for (int i = bl + 1; i <= rm; i++) {
    mx = max(mx, val[i]);
    ans = max(ans, val[bl] ^ mx);
  }
  solve(l, bl - 1);
  solve(bl + 1, r);
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = (1); i < (n + 1); i++) {
      scanf("%d", &val[i]);
    }
    initRMQ();
    ans = 0;
    solve(1, n);
    printf("%d\n", ans);
  }
  return 0;
}
