#include <bits/stdc++.h>
int main() {
  std::ios::sync_with_stdio(false);
  int n, k, x;
  std::cin >> n >> k >> x;
  long long arr[n];
  for (int i = 0; i < n; i++) std::cin >> arr[i];
  long long prefix[n + 2];
  long long suffix[n + 2];
  prefix[0] = 0;
  suffix[n + 1] = 0;
  for (int i = 1; i <= n; i++) {
    prefix[i] = prefix[i - 1] | arr[i - 1];
  }
  for (int i = n; i >= 1; i--) suffix[i] = suffix[i + 1] | arr[i - 1];
  long long mask = -1;
  long long tmp = 1;
  while (k-- > 0) tmp *= x;
  for (int i = 1; i <= n; i++) {
    long long t = prefix[i - 1] | (arr[i - 1] * tmp) | suffix[i + 1];
    if (t > mask) mask = t;
  }
  std::cout << mask << std::endl;
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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>
int main() {
  std::ios::sync_with_stdio(false);
  int n, k, x;
  std::cin >> n >> k >> x;
  long long arr[n];
  for (int i = 0; i < n; i++) std::cin >> arr[i];
  long long prefix[n + 2];
  long long suffix[n + 2];
  prefix[0] = 0;
  suffix[n + 1] = 0;
  for (int i = 1; i <= n; i++) {
    prefix[i] = prefix[i - 1] | arr[i - 1];
  }
  for (int i = n; i >= 1; i--) suffix[i] = suffix[i + 1] | arr[i - 1];
  long long mask = -1;
  long long tmp = 1;
  while (k-- > 0) tmp *= x;
  for (int i = 1; i <= n; i++) {
    long long t = prefix[i - 1] | (arr[i - 1] * tmp) | suffix[i + 1];
    if (t > mask) mask = t;
  }
  std::cout << mask << std::endl;
  return 0;
}

```