#include <bits/stdc++.h>
using namespace std;
int a[100005];
int ll[100005];
int rr[100005];
int L[100005][40];
int R[100005][40];
int zeros;
int n;
bool someCanSurvive;
void Print() {
  for (int bit = 31; bit >= 0; bit--) {
    printf("L: ");
    for (int i = 1; i <= n; i++) printf("%d ", L[i][bit]);
    printf("\nR: ");
    for (int i = 1; i <= n; i++) printf("%d ", R[i][bit]);
    printf("\n");
  }
}
int best;
int maks2;
int main() {
  best = 0;
  scanf("%d", &n);
  if (n == 1) {
    printf("0\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int bit = 31; bit >= 0; bit--) {
    zeros = 0;
    for (int i = 1; i <= n; i++) {
      if ((a[i] & (1 << bit)) != 0) {
        L[i][bit] = zeros;
        zeros = 0;
      } else {
        zeros++;
        L[i][bit] = 0;
      }
    }
    zeros = 0;
    for (int i = n; i >= 1; i--) {
      if ((a[i] & (1 << bit)) != 0) {
        R[i][bit] = zeros;
        zeros = 0;
      } else {
        zeros++;
        R[i][bit] = 0;
      }
    }
  }
  for (int i = 1; i <= n; i++) ll[i] = rr[i] = 1000000000;
  for (int bit = 31; bit >= 0; bit--) {
    for (int i = 1; i <= n; i++) {
      if (L[i][bit] > 0 || R[i][bit] > 0) {
        someCanSurvive = true;
        break;
      }
    }
    if (!someCanSurvive) continue;
    int i = 1;
    while (i <= n) {
      if (R[i][bit] > 0) {
        maks2 = a[i + 1];
        for (int j = 1; j <= R[i][bit]; j++) {
          maks2 = max(maks2, a[i + j]);
          best = max(best, a[i] ^ maks2);
        }
        i = i + R[i][bit] + 1;
      } else
        i++;
    }
    i = n;
    while (i >= 1) {
      if (L[i][bit] > 0) {
        maks2 = a[i - 1];
        for (int j = 1; j <= L[i][bit]; j++) {
          maks2 = max(maks2, a[i - j]);
          best = max(best, a[i] ^ maks2);
        }
        i = i - L[i][bit] - 1;
      } else
        i--;
    }
    printf("%d\n", best);
    return 0;
  }
  return 0;
}

### Prompt

Generate a CPP solution 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;
int a[100005];
int ll[100005];
int rr[100005];
int L[100005][40];
int R[100005][40];
int zeros;
int n;
bool someCanSurvive;
void Print() {
  for (int bit = 31; bit >= 0; bit--) {
    printf("L: ");
    for (int i = 1; i <= n; i++) printf("%d ", L[i][bit]);
    printf("\nR: ");
    for (int i = 1; i <= n; i++) printf("%d ", R[i][bit]);
    printf("\n");
  }
}
int best;
int maks2;
int main() {
  best = 0;
  scanf("%d", &n);
  if (n == 1) {
    printf("0\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int bit = 31; bit >= 0; bit--) {
    zeros = 0;
    for (int i = 1; i <= n; i++) {
      if ((a[i] & (1 << bit)) != 0) {
        L[i][bit] = zeros;
        zeros = 0;
      } else {
        zeros++;
        L[i][bit] = 0;
      }
    }
    zeros = 0;
    for (int i = n; i >= 1; i--) {
      if ((a[i] & (1 << bit)) != 0) {
        R[i][bit] = zeros;
        zeros = 0;
      } else {
        zeros++;
        R[i][bit] = 0;
      }
    }
  }
  for (int i = 1; i <= n; i++) ll[i] = rr[i] = 1000000000;
  for (int bit = 31; bit >= 0; bit--) {
    for (int i = 1; i <= n; i++) {
      if (L[i][bit] > 0 || R[i][bit] > 0) {
        someCanSurvive = true;
        break;
      }
    }
    if (!someCanSurvive) continue;
    int i = 1;
    while (i <= n) {
      if (R[i][bit] > 0) {
        maks2 = a[i + 1];
        for (int j = 1; j <= R[i][bit]; j++) {
          maks2 = max(maks2, a[i + j]);
          best = max(best, a[i] ^ maks2);
        }
        i = i + R[i][bit] + 1;
      } else
        i++;
    }
    i = n;
    while (i >= 1) {
      if (L[i][bit] > 0) {
        maks2 = a[i - 1];
        for (int j = 1; j <= L[i][bit]; j++) {
          maks2 = max(maks2, a[i - j]);
          best = max(best, a[i] ^ maks2);
        }
        i = i - L[i][bit] - 1;
      } else
        i--;
    }
    printf("%d\n", best);
    return 0;
  }
  return 0;
}

```