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;
struct node {
  int x, y;
  node(){};
  node(int x, int y) : x(x), y(y){};
} p[100005];
int tot;
bool v[33][100005];
int cnt[33];
int a[100005];
int main() {
  int i, j, n;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    int p = a[i], l = 0;
    while (p) {
      v[l][i] = p & 1;
      if (v[l][i]) cnt[l]++;
      l++;
      p >>= 1;
    }
  }
  int l = 32;
  while (!cnt[l] || cnt[l] == n) l--;
  if (l == -1) {
    puts("0");
    return 0;
  }
  p[tot++] = node(-1, -1);
  for (i = 0; i < n; i++)
    if (v[l][i] == 1) {
      j = i;
      while (v[l][j + 1] == 1 && j + 1 < n) j++;
      p[tot++] = node(i, j);
      i = j;
    }
  p[tot++] = node(n, n);
  int ans = 0;
  for (i = 1; i < tot - 1; i++) {
    int max = 0;
    for (j = p[i].x - 1; j >= p[i - 1].y + 1; j--) {
      if (max < a[j]) max = a[j];
      if ((max ^ a[p[i].x]) > ans) ans = max ^ a[p[i].x];
    }
    max = 0;
    for (j = p[i].y + 1; j <= p[i + 1].x - 1; j++) {
      if (max < a[j]) max = a[j];
      if ((max ^ a[p[i].y]) > ans) ans = max ^ a[p[i].y];
    }
  }
  printf("%d\n", ans);
  return 0;
}
