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;
int n;
int q[100010];
int a[100010];
int num[2][50];
int h;
void sep(int x, int id) {
  h = 0;
  while (x) {
    num[id][h++] = x % 2;
    x /= 2;
  }
}
void solve() {
  int ans = a[0] ^ a[1];
  int r = -1;
  q[++r] = a[0];
  if (a[1] >= a[0])
    q[0] = a[1];
  else
    q[++r] = a[1];
  for (int i = 2; i < n; ++i) {
    while (r && a[i] >= q[r]) {
      ans = max(ans, q[r] ^ a[i]);
      r--;
    }
    if (r != -1) ans = max(ans, q[r] ^ a[i]);
    q[++r] = a[i];
  }
  printf("%d\n", ans);
}
int main() {
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
    solve();
  }
  return 0;
}
