#include <bits/stdc++.h>
int n;
int s[100002];
int seg[560000];
void add(int lf, int rg, int id, int plc) {
  if (lf > plc || rg < plc) return;
  if (lf == plc && rg == plc) {
    seg[id] = s[plc];
    return;
  }
  int hf = (lf + rg) / 2;
  add(lf, hf, id + id + 1, plc);
  add(hf + 1, rg, id + id + 2, plc);
  seg[id] = seg[id + id + 1];
  if (seg[id] < seg[id + id + 2]) seg[id] = seg[id + id + 2];
}
int rmq_lf(int lf, int rg, int id, int krg) {
  if (krg < lf) return -1;
  if (rg <= krg)
    if (seg[id] < s[krg + 1]) return -1;
  if (lf == rg) {
    if (seg[id] > s[krg + 1]) return lf;
    return -1;
  }
  int hf = (lf + rg) / 2;
  int ret2 = rmq_lf(hf + 1, rg, id + id + 2, krg);
  if (ret2 > 0) return ret2;
  int ret1 = rmq_lf(lf, hf, id + id + 1, krg);
  return ret1;
}
int rmq_rg(int lf, int rg, int id, int klf) {
  if (klf > rg) return -1;
  if (lf >= klf)
    if (seg[id] < s[klf - 1]) return -1;
  if (lf == rg) {
    if (seg[id] > s[klf - 1]) return lf;
    return -1;
  }
  int hf = (lf + rg) / 2;
  int ret1 = rmq_rg(lf, hf, id + id + 1, klf);
  if (ret1 > 0) return ret1;
  int ret2 = rmq_rg(hf + 1, rg, id + id + 2, klf);
  return ret2;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &s[i]);
    add(1, n, 0, i);
  }
  int sol = 0;
  for (int i = 1; i <= n; i++) {
    if (i > 1) {
      int value = s[rmq_lf(1, n, 0, i - 1)];
      if (value > s[i]) {
        if (sol < (value ^ s[i])) {
          sol = (value ^ s[i]);
        }
      }
    }
    if (i < n) {
      int value = s[rmq_rg(1, n, 0, i + 1)];
      if (value > s[i]) {
        if (sol < (value ^ s[i])) {
          sol = (value ^ s[i]);
        }
      }
    }
  }
  printf("%d\n", sol);
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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}.

### Response

```cpp
#include <bits/stdc++.h>
int n;
int s[100002];
int seg[560000];
void add(int lf, int rg, int id, int plc) {
  if (lf > plc || rg < plc) return;
  if (lf == plc && rg == plc) {
    seg[id] = s[plc];
    return;
  }
  int hf = (lf + rg) / 2;
  add(lf, hf, id + id + 1, plc);
  add(hf + 1, rg, id + id + 2, plc);
  seg[id] = seg[id + id + 1];
  if (seg[id] < seg[id + id + 2]) seg[id] = seg[id + id + 2];
}
int rmq_lf(int lf, int rg, int id, int krg) {
  if (krg < lf) return -1;
  if (rg <= krg)
    if (seg[id] < s[krg + 1]) return -1;
  if (lf == rg) {
    if (seg[id] > s[krg + 1]) return lf;
    return -1;
  }
  int hf = (lf + rg) / 2;
  int ret2 = rmq_lf(hf + 1, rg, id + id + 2, krg);
  if (ret2 > 0) return ret2;
  int ret1 = rmq_lf(lf, hf, id + id + 1, krg);
  return ret1;
}
int rmq_rg(int lf, int rg, int id, int klf) {
  if (klf > rg) return -1;
  if (lf >= klf)
    if (seg[id] < s[klf - 1]) return -1;
  if (lf == rg) {
    if (seg[id] > s[klf - 1]) return lf;
    return -1;
  }
  int hf = (lf + rg) / 2;
  int ret1 = rmq_rg(lf, hf, id + id + 1, klf);
  if (ret1 > 0) return ret1;
  int ret2 = rmq_rg(hf + 1, rg, id + id + 2, klf);
  return ret2;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &s[i]);
    add(1, n, 0, i);
  }
  int sol = 0;
  for (int i = 1; i <= n; i++) {
    if (i > 1) {
      int value = s[rmq_lf(1, n, 0, i - 1)];
      if (value > s[i]) {
        if (sol < (value ^ s[i])) {
          sol = (value ^ s[i]);
        }
      }
    }
    if (i < n) {
      int value = s[rmq_rg(1, n, 0, i + 1)];
      if (value > s[i]) {
        if (sol < (value ^ s[i])) {
          sol = (value ^ s[i]);
        }
      }
    }
  }
  printf("%d\n", sol);
  return 0;
}

```