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>
#pragma GCC optimize(2)
using namespace std;
inline void swap(int &x, int &y) { x ^= y ^= x ^= y; }
const double pi = acos(-1.0);
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const long long inf = 0x7f7f7f7f;
const int mod = 1e9 + 7;
const int N = 2e5 + 5;
int n, m, t;
int i, j, k;
long long a[N];
stack<int> s;
int main() {
  while (scanf("%d", &n) == 1) {
    for (i = 1; i <= n; i++) scanf("%d", &a[i]);
    long long ans = 0;
    for (i = 1; i <= n; i++) {
      while (s.size() && a[s.top()] <= a[i]) s.pop();
      if (s.size()) ans = max(ans, a[i] ^ a[s.top()]);
      s.push(i);
    }
    stack<int> hollow;
    swap(hollow, s);
    for (i = n; i; i--) {
      while (s.size() && a[s.top()] <= a[i]) s.pop();
      if (s.size()) ans = max(ans, a[i] ^ a[s.top()]);
      s.push(i);
    }
    printf("%lld\n", ans);
  }
}
