#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k, x, p = 1;
  cin >> n >> k >> x;
  vector<long long> a(n + 1), orstg(n + 1), ordrp(n + 1);
  for (int i = 0; i < n; i++) cin >> a[i];
  orstg[0] = a[0];
  for (int i = 1; i < n; i++) orstg[i] = orstg[i - 1] | a[i];
  ordrp[n - 1] = a[n - 1];
  for (int i = n - 2; i >= 0; i--) ordrp[i] = ordrp[i + 1] | a[i];
  while (k--) p *= x;
  if (n == 1) {
    cout << a[0] * p << "\n";
    return 0;
  }
  long long ans = max((a[0] * p) | ordrp[1], orstg[n - 2] | (a[n - 1] * p));
  for (int i = 1; i < n - 1; i++)
    ans = max(ans, orstg[i - 1] | ordrp[i + 1] | (a[i] * p));
  cout << ans << "\n";
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
int main() {
  int n, k, x, p = 1;
  cin >> n >> k >> x;
  vector<long long> a(n + 1), orstg(n + 1), ordrp(n + 1);
  for (int i = 0; i < n; i++) cin >> a[i];
  orstg[0] = a[0];
  for (int i = 1; i < n; i++) orstg[i] = orstg[i - 1] | a[i];
  ordrp[n - 1] = a[n - 1];
  for (int i = n - 2; i >= 0; i--) ordrp[i] = ordrp[i + 1] | a[i];
  while (k--) p *= x;
  if (n == 1) {
    cout << a[0] * p << "\n";
    return 0;
  }
  long long ans = max((a[0] * p) | ordrp[1], orstg[n - 2] | (a[n - 1] * p));
  for (int i = 1; i < n - 1; i++)
    ans = max(ans, orstg[i - 1] | ordrp[i + 1] | (a[i] * p));
  cout << ans << "\n";
  return 0;
}

```