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, a[1000 * 100 + 100], t, start, Max;
bool check() {
  for (int i = 1; i <= n; i++)
    if (a[i] < t) return false;
  return true;
}
bool mark[1000 * 100 + 100];
void calc(int x, int y) {
  int tmax = 0;
  if (a[x] != 0)
    for (int i = x + 1; i < y; i++) {
      tmax = max(tmax, a[i]);
      Max = max(Max, tmax ^ a[x]);
    }
  tmax = 0;
  if (a[y] != 0)
    for (int i = y - 1; i > x; i--) {
      tmax = max(tmax, a[i]);
      Max = max(Max, tmax ^ a[y]);
    }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  while (check()) {
    for (int i = 1; i <= n; i++) a[i] -= t;
    t = 1;
    for (int i = 1; i <= n; i++) {
      while (a[i] >= 2 * t) t *= 2;
    }
  }
  mark[0] = true;
  mark[n + 1] = true;
  for (int i = 1; i <= n; i++)
    if (a[i] >= t) mark[i] = true;
  for (int i = 1; i <= n + 1; i++)
    if (mark[i]) {
      calc(start, i);
      start = i;
    }
  cout << Max;
}
