#include <bits/stdc++.h>
using namespace std;
map<int, pair<int, int>, std::greater<int> > order;
int main() {
  long long n, k, x;
  cin >> n >> k >> x;
  long long arr[n + 1];
  long long LOR[n + 2];
  long long ROR[n + 2];
  LOR[0] = 0;
  ROR[n + 1] = 0;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    LOR[i] = LOR[i - 1] | arr[i];
  }
  for (int i = n; i >= 1; i--) {
    ROR[i] = ROR[i + 1] | arr[i];
  }
  ROR[0] = ROR[1];
  LOR[n + 1] = LOR[n];
  long long ans = 0;
  long long mul = 1;
  while (k--) {
    mul = mul * x;
  }
  for (int i = 1; i <= n; i++) {
    ans = max(ans, LOR[i - 1] | mul * arr[i] | ROR[i + 1]);
  }
  cout << ans;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
map<int, pair<int, int>, std::greater<int> > order;
int main() {
  long long n, k, x;
  cin >> n >> k >> x;
  long long arr[n + 1];
  long long LOR[n + 2];
  long long ROR[n + 2];
  LOR[0] = 0;
  ROR[n + 1] = 0;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    LOR[i] = LOR[i - 1] | arr[i];
  }
  for (int i = n; i >= 1; i--) {
    ROR[i] = ROR[i + 1] | arr[i];
  }
  ROR[0] = ROR[1];
  LOR[n + 1] = LOR[n];
  long long ans = 0;
  long long mul = 1;
  while (k--) {
    mul = mul * x;
  }
  for (int i = 1; i <= n; i++) {
    ans = max(ans, LOR[i - 1] | mul * arr[i] | ROR[i + 1]);
  }
  cout << ans;
  return 0;
}

```