#include <bits/stdc++.h>
using namespace std;
const int N = 200000;
long long A[N + 2], L[N + 2], R[N + 2];
bool solve(int testNumber) {
  int n, k, x;
  if (scanf("%d %d %d", &n, &k, &x) == EOF) return false;
  R[n + 1] = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &A[i]);
    L[i] = L[i - 1] | A[i];
  }
  for (int i = n; i >= 0; i--) R[i] = R[i + 1] | A[i];
  long long ret = 0;
  for (int i = 1; i <= n; i++) {
    long long aux = A[i];
    for (int j = 0; j < k; j++) aux *= x;
    ret = max(ret, L[i - 1] | R[i + 1] | aux);
  }
  printf("%lld\n", ret);
  return true;
}
int main() {
  for (int i = 1; solve(i); i++)
    ;
  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 N = 200000;
long long A[N + 2], L[N + 2], R[N + 2];
bool solve(int testNumber) {
  int n, k, x;
  if (scanf("%d %d %d", &n, &k, &x) == EOF) return false;
  R[n + 1] = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &A[i]);
    L[i] = L[i - 1] | A[i];
  }
  for (int i = n; i >= 0; i--) R[i] = R[i + 1] | A[i];
  long long ret = 0;
  for (int i = 1; i <= n; i++) {
    long long aux = A[i];
    for (int j = 0; j < k; j++) aux *= x;
    ret = max(ret, L[i - 1] | R[i + 1] | aux);
  }
  printf("%lld\n", ret);
  return true;
}
int main() {
  for (int i = 1; solve(i); i++)
    ;
  return 0;
}

```