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 inf = 0x3f3f3f3f;
const int maxn = 100050;
int n, a[maxn];
int s[maxn], top;
inline int read() {
  int x = 0, t = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') t = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * t;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    while (top && a[s[top]] < a[i]) top--;
    if (top) ans = max(ans, a[i] ^ a[s[top]]);
    s[++top] = i;
  }
  reverse(a + 1, a + n + 1);
  top = 0;
  memset(s, 0, sizeof(s));
  for (int i = 1; i <= n; i++) {
    while (top && a[s[top]] < a[i]) top--;
    if (top) ans = max(ans, a[i] ^ a[s[top]]);
    s[++top] = i;
  }
  cout << ans << endl;
  return 0;
}
