#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
const int N = 100100;
long long dp[N][20];
int LOG[N];
long long arr[N];
void initLOG() {
  LOG[0] = -1;
  for (int i = 1; i <= 100000; ++i) {
    LOG[i] = LOG[i - 1];
    LOG[i] += !(i & (i - 1));
  }
}
void initRMQ(int n) {
  for (int i = 1; i <= n; ++i) {
    dp[i][0] = arr[i];
  }
  for (int j = 1; j <= LOG[n]; ++j) {
    for (int i = 1; i + (1 << j) - 1 <= n; ++i) {
      dp[i][j] = max(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
    }
  }
}
long long ST(int l, int r) {
  int k = LOG[r - l + 1];
  return max(dp[l][k], dp[r - (1 << k) + 1][k]);
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    long long ans = 0;
    for (int i = 1; i <= n; ++i) {
      scanf("%I64d", &arr[i]);
    }
    initLOG();
    initRMQ(n);
    for (int i = 1; i <= n; ++i) {
      int l = i + 1, r = n, mid;
      long long tmp = arr[i];
      while (l <= r) {
        mid = (l + r) >> 1;
        long long maxs = ST(i + 1, mid);
        if (maxs >= arr[i]) {
          tmp = maxs;
          r = mid - 1;
        } else {
          l = mid + 1;
        }
      }
      ans = max(ans, tmp ^ arr[i]);
      l = 1;
      r = i - 1;
      tmp = arr[i];
      while (l <= r) {
        mid = (l + r) >> 1;
        long long maxs = ST(mid, i - 1);
        if (maxs > arr[i]) {
          tmp = maxs;
          l = mid + 1;
        } else {
          r = mid - 1;
        }
      }
      ans = max(ans, tmp ^ arr[i]);
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

### 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;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
const int N = 100100;
long long dp[N][20];
int LOG[N];
long long arr[N];
void initLOG() {
  LOG[0] = -1;
  for (int i = 1; i <= 100000; ++i) {
    LOG[i] = LOG[i - 1];
    LOG[i] += !(i & (i - 1));
  }
}
void initRMQ(int n) {
  for (int i = 1; i <= n; ++i) {
    dp[i][0] = arr[i];
  }
  for (int j = 1; j <= LOG[n]; ++j) {
    for (int i = 1; i + (1 << j) - 1 <= n; ++i) {
      dp[i][j] = max(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
    }
  }
}
long long ST(int l, int r) {
  int k = LOG[r - l + 1];
  return max(dp[l][k], dp[r - (1 << k) + 1][k]);
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    long long ans = 0;
    for (int i = 1; i <= n; ++i) {
      scanf("%I64d", &arr[i]);
    }
    initLOG();
    initRMQ(n);
    for (int i = 1; i <= n; ++i) {
      int l = i + 1, r = n, mid;
      long long tmp = arr[i];
      while (l <= r) {
        mid = (l + r) >> 1;
        long long maxs = ST(i + 1, mid);
        if (maxs >= arr[i]) {
          tmp = maxs;
          r = mid - 1;
        } else {
          l = mid + 1;
        }
      }
      ans = max(ans, tmp ^ arr[i]);
      l = 1;
      r = i - 1;
      tmp = arr[i];
      while (l <= r) {
        mid = (l + r) >> 1;
        long long maxs = ST(mid, i - 1);
        if (maxs > arr[i]) {
          tmp = maxs;
          l = mid + 1;
        } else {
          r = mid - 1;
        }
      }
      ans = max(ans, tmp ^ arr[i]);
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```