#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int N, K, X;
  cin >> N >> K >> X;
  long long int Y = 1;
  int A = 0;
  long long int arr[N], AL[N + 1], AR[N + 1];
  long long int maxi = 0;
  AL[0] = 0;
  AR[N] = 0;
  while (K--) Y *= X;
  for (int i = 0; i < N; i++) {
    cin >> arr[i];
    AL[i + 1] = (AL[i] | arr[i]);
  }
  for (int i = N - 1; i >= 0; i--) {
    AR[i] = (AR[i + 1] | arr[i]);
  }
  for (int i = 0; i < N; i++) {
    maxi = max(maxi, ((AL[i] | (arr[i] * Y)) | AR[i + 1]));
  }
  cout << maxi << endl;
}

### Prompt

Please formulate 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;
int main() {
  long long int N, K, X;
  cin >> N >> K >> X;
  long long int Y = 1;
  int A = 0;
  long long int arr[N], AL[N + 1], AR[N + 1];
  long long int maxi = 0;
  AL[0] = 0;
  AR[N] = 0;
  while (K--) Y *= X;
  for (int i = 0; i < N; i++) {
    cin >> arr[i];
    AL[i + 1] = (AL[i] | arr[i]);
  }
  for (int i = N - 1; i >= 0; i--) {
    AR[i] = (AR[i + 1] | arr[i]);
  }
  for (int i = 0; i < N; i++) {
    maxi = max(maxi, ((AL[i] | (arr[i] * Y)) | AR[i + 1]));
  }
  cout << maxi << endl;
}

```