#include <bits/stdc++.h>
using namespace std;
const int MAXN = 111111;
int T[MAXN * 4], a[MAXN], n;
void build(int x, int l, int r) {
  if (l == r)
    T[x] = a[l];
  else {
    int mid = (l + r) >> 1;
    build((x << 1), l, mid);
    build((((x << 1)) + 1), mid + 1, r);
    T[x] = max(T[(x << 1)], T[(((x << 1)) + 1)]);
  }
}
int query(int x, int l, int r, int ll, int rr) {
  if (ll <= l && rr >= r) {
    return T[x];
  } else {
    int mid = (l + r) >> 1;
    int ret = 0;
    if (ll <= mid) ret = max(ret, query((x << 1), l, mid, ll, rr));
    if (rr > mid) ret = max(ret, query((((x << 1)) + 1), mid + 1, r, ll, rr));
    return ret;
  }
}
int getnum(int pos, int x) {
  int l = pos, r = n;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (query(1, 1, n, pos, mid) >= x) {
      r = mid;
    } else
      l = mid + 1;
  }
  while (query(1, 1, n, pos, l) < x && l <= n) l++;
  if (l > n) return -1;
  return a[l];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  build(1, 1, n);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int t = getnum(i + 1, a[i] + 1);
    if (t < 0) continue;
    ans = max(a[i] ^ t, ans);
  }
  reverse(a + 1, a + n + 1);
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    int t = getnum(i + 1, a[i] + 1);
    if (t < 0) continue;
    ans = max(a[i] ^ t, ans);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp to 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>
using namespace std;
const int MAXN = 111111;
int T[MAXN * 4], a[MAXN], n;
void build(int x, int l, int r) {
  if (l == r)
    T[x] = a[l];
  else {
    int mid = (l + r) >> 1;
    build((x << 1), l, mid);
    build((((x << 1)) + 1), mid + 1, r);
    T[x] = max(T[(x << 1)], T[(((x << 1)) + 1)]);
  }
}
int query(int x, int l, int r, int ll, int rr) {
  if (ll <= l && rr >= r) {
    return T[x];
  } else {
    int mid = (l + r) >> 1;
    int ret = 0;
    if (ll <= mid) ret = max(ret, query((x << 1), l, mid, ll, rr));
    if (rr > mid) ret = max(ret, query((((x << 1)) + 1), mid + 1, r, ll, rr));
    return ret;
  }
}
int getnum(int pos, int x) {
  int l = pos, r = n;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (query(1, 1, n, pos, mid) >= x) {
      r = mid;
    } else
      l = mid + 1;
  }
  while (query(1, 1, n, pos, l) < x && l <= n) l++;
  if (l > n) return -1;
  return a[l];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  build(1, 1, n);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int t = getnum(i + 1, a[i] + 1);
    if (t < 0) continue;
    ans = max(a[i] ^ t, ans);
  }
  reverse(a + 1, a + n + 1);
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    int t = getnum(i + 1, a[i] + 1);
    if (t < 0) continue;
    ans = max(a[i] ^ t, ans);
  }
  cout << ans << endl;
  return 0;
}

```