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 double eps = 1e-6;
const int mod = 998244353;
const int maxn = 2e6 + 100;
const int maxm = 2e6 + 100;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
int n, m;
int a[maxn];
int l[maxn], r[maxn];
int main() {
  scanf("%d", &n);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    l[i] = r[i] = i;
  }
  a[0] = a[n + 1] = inf;
  for (int i = 1; i <= n; i++) {
    while (a[i] >= a[l[i] - 1]) l[i] = l[l[i] - 1];
  }
  for (int i = n; i >= 1; i--) {
    while (a[i] >= a[r[i] + 1]) r[i] = r[r[i] + 1];
  }
  for (int i = 1; i <= n; i++) {
    int L = l[i];
    int R = r[i];
    if (L != 1) ans = max(ans, a[i] ^ a[L - 1]);
    if (R != n) ans = max(ans, a[i] ^ a[R + 1]);
  }
  printf("%d", ans);
  return 0;
}
