#include <bits/stdc++.h>
using namespace std;
struct node {
  int l, r, maxx;
};
node g[800000];
int X, ll, rr, mid, i, j, N, ans;
int A[100005];
void build(int a, int b, int u) {
  if (a == b) {
    g[u] = (node){a, b, A[a]};
    return;
  }
  g[u] = (node){a, b, A[a]};
  int mid = (a + b) >> 1;
  build(a, mid, 2 * u);
  build(mid + 1, b, 2 * u + 1);
  g[u].maxx = max(g[2 * u].maxx, g[2 * u + 1].maxx);
}
int getmax(int a, int b, int u) {
  if (a > b) return -1000000000;
  int left = g[u].l;
  int right = g[u].r;
  if (a > right || left > b) return -1000000000;
  if (a <= left && right <= b) return g[u].maxx;
  return max(getmax(a, b, 2 * u), getmax(a, b, 2 * u + 1));
}
int main() {
  scanf("%d", &N);
  for (i = 1; i <= N; i++) scanf("%d", &A[i]);
  build(1, N, 1);
  for (i = 1; i <= N; i++) {
    if (getmax(1, i - 1, 1) > A[i]) {
      ll = 1;
      rr = i - 1;
      while (ll < rr) {
        mid = (ll + rr) >> 1;
        if (getmax(mid, i - 1, 1) > A[i])
          ll = mid;
        else
          rr = mid - 1;
        if (rr == ll + 1)
          if (getmax(rr, i - 1, 1) > A[i])
            ll++;
          else
            rr--;
      }
      X = rr;
      ans = max(ans, A[i] ^ A[X]);
    }
    if (getmax(i + 1, N, 1) > A[i]) {
      ll = i + 1;
      rr = N;
      while (ll < rr) {
        mid = (ll + rr) >> 1;
        if (getmax(i + 1, mid, 1) > A[i])
          rr = mid;
        else
          ll = mid + 1;
        if (ll == rr - 1)
          if (getmax(i + 1, ll, 1) > A[i])
            rr--;
          else
            ll++;
      }
      X = rr;
      ans = max(ans, A[i] ^ A[X]);
    }
  }
  printf("%d\n", ans);
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
using namespace std;
struct node {
  int l, r, maxx;
};
node g[800000];
int X, ll, rr, mid, i, j, N, ans;
int A[100005];
void build(int a, int b, int u) {
  if (a == b) {
    g[u] = (node){a, b, A[a]};
    return;
  }
  g[u] = (node){a, b, A[a]};
  int mid = (a + b) >> 1;
  build(a, mid, 2 * u);
  build(mid + 1, b, 2 * u + 1);
  g[u].maxx = max(g[2 * u].maxx, g[2 * u + 1].maxx);
}
int getmax(int a, int b, int u) {
  if (a > b) return -1000000000;
  int left = g[u].l;
  int right = g[u].r;
  if (a > right || left > b) return -1000000000;
  if (a <= left && right <= b) return g[u].maxx;
  return max(getmax(a, b, 2 * u), getmax(a, b, 2 * u + 1));
}
int main() {
  scanf("%d", &N);
  for (i = 1; i <= N; i++) scanf("%d", &A[i]);
  build(1, N, 1);
  for (i = 1; i <= N; i++) {
    if (getmax(1, i - 1, 1) > A[i]) {
      ll = 1;
      rr = i - 1;
      while (ll < rr) {
        mid = (ll + rr) >> 1;
        if (getmax(mid, i - 1, 1) > A[i])
          ll = mid;
        else
          rr = mid - 1;
        if (rr == ll + 1)
          if (getmax(rr, i - 1, 1) > A[i])
            ll++;
          else
            rr--;
      }
      X = rr;
      ans = max(ans, A[i] ^ A[X]);
    }
    if (getmax(i + 1, N, 1) > A[i]) {
      ll = i + 1;
      rr = N;
      while (ll < rr) {
        mid = (ll + rr) >> 1;
        if (getmax(i + 1, mid, 1) > A[i])
          rr = mid;
        else
          ll = mid + 1;
        if (ll == rr - 1)
          if (getmax(i + 1, ll, 1) > A[i])
            rr--;
          else
            ll++;
      }
      X = rr;
      ans = max(ans, A[i] ^ A[X]);
    }
  }
  printf("%d\n", ans);
}

```