#include <bits/stdc++.h>
using namespace std;
unsigned long long n, k, x, l[200100];
unsigned long long accl[200100], accr[200100];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k >> x;
  for (int i = 1; i <= n; i++) {
    cin >> l[i];
    accl[i] = accl[i - 1] | l[i];
  }
  for (int i = n; i > 0; i--) {
    accr[i] = accr[i + 1] | l[i];
  }
  unsigned long long pot = 1ull;
  for (int i = 1; i <= k; i++) pot *= x;
  unsigned long long sol = 0;
  for (int i = 1; i <= n; i++) {
    sol = max(sol, accl[i - 1] | (l[i] * pot) | accr[i + 1]);
  }
  cout << sol << endl;
}

### Prompt

In cpp, your task is to solve 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>
using namespace std;
unsigned long long n, k, x, l[200100];
unsigned long long accl[200100], accr[200100];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k >> x;
  for (int i = 1; i <= n; i++) {
    cin >> l[i];
    accl[i] = accl[i - 1] | l[i];
  }
  for (int i = n; i > 0; i--) {
    accr[i] = accr[i + 1] | l[i];
  }
  unsigned long long pot = 1ull;
  for (int i = 1; i <= k; i++) pot *= x;
  unsigned long long sol = 0;
  for (int i = 1; i <= n; i++) {
    sol = max(sol, accl[i - 1] | (l[i] * pot) | accr[i + 1]);
  }
  cout << sol << endl;
}

```