#include <bits/stdc++.h>
using namespace std;
unsigned long long a[200203];
unsigned long long b[203420];
unsigned long long c[203433];
int main() {
  unsigned long long n, m, x;
  cin >> n >> m >> x;
  unsigned long long i, j, k;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (i = 1; i <= n; i++) {
    b[i] = b[i - 1] | a[i];
  }
  for (j = n; j >= 1; j--) {
    c[j] = c[j + 1] | a[j];
  }
  unsigned long long tmp;
  unsigned long long mx = 0;
  int z = 1;
  for (i = 0; i <= m; i++) {
    for (j = 1; j <= n; j++) {
      tmp = b[j - 1] | (a[j] * z) | c[j + 1];
      mx = max(mx, tmp);
    }
    z *= x;
  }
  cout << mx << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

You are given n numbers a1, a2, ..., an. You can perform at most k operations. For each operation you can multiply one of the numbers by x. We want to make <image> as large as possible, where <image> denotes the bitwise OR. 

Find the maximum possible value of <image> after performing at most k operations optimally.

Input

The first line contains three integers n, k and x (1 ≤ n ≤ 200 000, 1 ≤ k ≤ 10, 2 ≤ x ≤ 8).

The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).

Output

Output the maximum value of a bitwise OR of sequence elements after performing operations.

Examples

Input

3 1 2
1 1 1


Output

3


Input

4 2 3
1 2 4 8


Output

79

Note

For the first sample, any possible choice of doing one operation will result the same three numbers 1, 1, 2 so the result is <image>. 

For the second sample if we multiply 8 by 3 two times we'll get 72. In this case the numbers will become 1, 2, 4, 72 so the OR value will be 79 and is the largest possible result.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
unsigned long long a[200203];
unsigned long long b[203420];
unsigned long long c[203433];
int main() {
  unsigned long long n, m, x;
  cin >> n >> m >> x;
  unsigned long long i, j, k;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (i = 1; i <= n; i++) {
    b[i] = b[i - 1] | a[i];
  }
  for (j = n; j >= 1; j--) {
    c[j] = c[j + 1] | a[j];
  }
  unsigned long long tmp;
  unsigned long long mx = 0;
  int z = 1;
  for (i = 0; i <= m; i++) {
    for (j = 1; j <= n; j++) {
      tmp = b[j - 1] | (a[j] * z) | c[j + 1];
      mx = max(mx, tmp);
    }
    z *= x;
  }
  cout << mx << endl;
  return 0;
}

```