#include <bits/stdc++.h>
using namespace std;
const int sz = 212345;
long long a[sz], pre[sz], suf[sz];
int main() {
  int n, k, x, i, j;
  long long ans, tmp = 1LL, ta;
  scanf("%d", &n);
  ;
  scanf("%d", &k);
  ;
  scanf("%d", &x);
  ;
  pre[0] = 0LL;
  for (i = 0; i < n; i++) {
    scanf("%I64d", &a[i]);
    ;
    pre[i + 1] = pre[i] | a[i];
  }
  suf[n + 1] = 0LL;
  for (i = n; i >= 1; i--) suf[i] = suf[i + 1] | a[i - 1];
  for (i = 0; i < k; i++) tmp *= (long long)x;
  ans = -1LL;
  for (i = 0; i < n; i++) {
    ta = a[i] * tmp;
    ta = ta | pre[i];
    ta |= suf[i + 2];
    if (ta > ans) ans = ta;
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Generate 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 sz = 212345;
long long a[sz], pre[sz], suf[sz];
int main() {
  int n, k, x, i, j;
  long long ans, tmp = 1LL, ta;
  scanf("%d", &n);
  ;
  scanf("%d", &k);
  ;
  scanf("%d", &x);
  ;
  pre[0] = 0LL;
  for (i = 0; i < n; i++) {
    scanf("%I64d", &a[i]);
    ;
    pre[i + 1] = pre[i] | a[i];
  }
  suf[n + 1] = 0LL;
  for (i = n; i >= 1; i--) suf[i] = suf[i + 1] | a[i - 1];
  for (i = 0; i < k; i++) tmp *= (long long)x;
  ans = -1LL;
  for (i = 0; i < n; i++) {
    ta = a[i] * tmp;
    ta = ta | pre[i];
    ta |= suf[i + 2];
    if (ta > ans) ans = ta;
  }
  printf("%I64d\n", ans);
  return 0;
}

```