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 = 2000000000;
const int N = 100010;
int low[5 * N], high[5 * N], Min[5 * N];
void Creat(int l, int r, int p) {
  low[p] = l, high[p] = r;
  if (l == r) {
    scanf("%d", &Min[p]);
    return;
  }
  int mid = (l + r) >> 1;
  Creat(l, mid, 2 * p);
  Creat(mid + 1, r, 2 * p + 1);
  Min[p] = min(Min[2 * p], Min[2 * p + 1]);
}
void Query(int p, int &pos, int &val) {
  if (low[p] == high[p]) {
    pos = low[p];
    val = Min[p];
    Min[p] = INF;
    return;
  }
  int mid = (low[p] + high[p]) >> 1;
  if (Min[2 * p] < Min[2 * p + 1])
    Query(2 * p, pos, val);
  else
    Query(2 * p + 1, pos, val);
  Min[p] = min(Min[2 * p], Min[2 * p + 1]);
}
int Left(int p, int l, int r) {
  if (Min[p] == INF) return -1;
  if (low[p] == high[p]) return Min[p];
  int mid = (low[p] + high[p]) >> 1;
  if (mid < l)
    return Left(2 * p + 1, l, r);
  else if (r <= mid)
    return Left(2 * p, l, r);
  else {
    int tmp = Left(2 * p, l, r);
    if (tmp != -1) return tmp;
    return Left(2 * p + 1, l, r);
  }
}
int Right(int p, int l, int r) {
  if (Min[p] == INF) return -1;
  if (low[p] == high[p]) return Min[p];
  int mid = (low[p] + high[p]) >> 1;
  if (mid < l)
    return Right(2 * p + 1, l, r);
  else if (r <= mid)
    return Right(2 * p, l, r);
  else {
    int tmp = Right(2 * p + 1, l, r);
    if (tmp != -1) return tmp;
    return Right(2 * p, l, r);
  }
}
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    int Max = 0;
    int pos, val;
    Creat(1, n, 1);
    for (int i = 1; i < n; i++) {
      Query(1, pos, val);
      if (pos + 1 <= n) {
        int tmp = Left(1, pos + 1, n);
        if (tmp != -1) Max = max(Max, (val ^ tmp));
      }
      if (1 <= pos - 1) {
        int tmp = Right(1, 1, pos - 1);
        if (tmp != -1) Max = max(Max, (val ^ tmp));
      }
    }
    printf("%d\n", Max);
  }
  return 0;
}
