#include <bits/stdc++.h>
using namespace std;
long long power(long long base, long long exp) {
  base %= 1000000007;
  long long result = 1;
  while (exp > 0) {
    if (exp & 1) result = (result * base) % 1000000007;
    base = (base * base) % 1000000007;
    exp >>= 1;
  }
  return result;
}
const int MAX = 200010;
int arr[MAX];
long long l[MAX];
long long r[MAX];
int main() {
  int n, k, x;
  long long ml = 1;
  scanf("%d", &n);
  scanf("%d", &k);
  scanf("%d", &x);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &arr[i]);
    l[i] = (l[i - 1] | arr[i]);
  }
  for (int i = n; i > 0; --i) r[i] = (r[i + 1] | arr[i]);
  long long ans = 0;
  while (k--) ml *= x;
  for (int i = 1; i <= n; ++i) {
    ans = max(ans, ml * arr[i] | l[i - 1] | r[i + 1]);
  }
  printf("%lld\n", ans);
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
long long power(long long base, long long exp) {
  base %= 1000000007;
  long long result = 1;
  while (exp > 0) {
    if (exp & 1) result = (result * base) % 1000000007;
    base = (base * base) % 1000000007;
    exp >>= 1;
  }
  return result;
}
const int MAX = 200010;
int arr[MAX];
long long l[MAX];
long long r[MAX];
int main() {
  int n, k, x;
  long long ml = 1;
  scanf("%d", &n);
  scanf("%d", &k);
  scanf("%d", &x);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &arr[i]);
    l[i] = (l[i - 1] | arr[i]);
  }
  for (int i = n; i > 0; --i) r[i] = (r[i + 1] | arr[i]);
  long long ans = 0;
  while (k--) ml *= x;
  for (int i = 1; i <= n; ++i) {
    ans = max(ans, ml * arr[i] | l[i - 1] | r[i + 1]);
  }
  printf("%lld\n", ans);
}

```