#include <bits/stdc++.h>
using namespace std;
const int oo = 2e6 + 22;
int N, k, x;
long long int ans, qQq;
long long int a[oo], foo[oo], soo[oo];
int main() {
  scanf("%d %d %d", &N, &k, &x);
  for (int i = 1; i <= N; i++) scanf("%lld", a + i);
  qQq = 1;
  while (k--) qQq *= x;
  foo[0] = 0;
  for (int i = 1; i <= N; i++) foo[i] = foo[i - 1] | a[i];
  soo[N + 1] = 0;
  for (int i = N; i > 0; i--) soo[i] = soo[i + 1] | a[i];
  ans = 0;
  for (int i = 1; i <= N; i++)
    ans = max(ans, foo[i - 1] | a[i] * qQq | soo[i + 1]);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to 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;
const int oo = 2e6 + 22;
int N, k, x;
long long int ans, qQq;
long long int a[oo], foo[oo], soo[oo];
int main() {
  scanf("%d %d %d", &N, &k, &x);
  for (int i = 1; i <= N; i++) scanf("%lld", a + i);
  qQq = 1;
  while (k--) qQq *= x;
  foo[0] = 0;
  for (int i = 1; i <= N; i++) foo[i] = foo[i - 1] | a[i];
  soo[N + 1] = 0;
  for (int i = N; i > 0; i--) soo[i] = soo[i + 1] | a[i];
  ans = 0;
  for (int i = 1; i <= N; i++)
    ans = max(ans, foo[i - 1] | a[i] * qQq | soo[i + 1]);
  printf("%lld\n", ans);
  return 0;
}

```