#include <bits/stdc++.h>
using namespace std;
void optimizeIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
int main() {
  optimizeIO();
  long long n, k, x;
  cin >> n >> k >> x;
  vector<long long> v(n);
  for (typeof(n) i = 0; i < n; i++) cin >> v[i];
  long long t = 1;
  while (k--) t *= x;
  vector<long long> pre(n);
  vector<long long> pos(n);
  if (n == 1) {
    cout << t * v[0] << endl;
  } else {
    pre[1] = v[0];
    pos[n - 2] = v[n - 1];
    for (typeof(n) i = 2; i < n; i++) {
      pre[i] = pre[i - 1] | v[i - 1];
    }
    for (typeof(n - 3) i = n - 3; i >= 0; i--) {
      pos[i] = pos[i + 1] | v[i + 1];
    }
    long long maxx = -1;
    for (typeof(n) i = 0; i < n; i++) {
      if (i == 0) {
        maxx = max(maxx, pos[0] | (t * v[i]));
        continue;
      } else if (i == n - 1) {
        maxx = max(maxx, pre[i] | (t * v[i]));
        break;
      }
      maxx = max(maxx, pre[i] | pos[i] | (t * v[i]));
    }
    cout << maxx << endl;
  }
  return 0;
}

### 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;
void optimizeIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
int main() {
  optimizeIO();
  long long n, k, x;
  cin >> n >> k >> x;
  vector<long long> v(n);
  for (typeof(n) i = 0; i < n; i++) cin >> v[i];
  long long t = 1;
  while (k--) t *= x;
  vector<long long> pre(n);
  vector<long long> pos(n);
  if (n == 1) {
    cout << t * v[0] << endl;
  } else {
    pre[1] = v[0];
    pos[n - 2] = v[n - 1];
    for (typeof(n) i = 2; i < n; i++) {
      pre[i] = pre[i - 1] | v[i - 1];
    }
    for (typeof(n - 3) i = n - 3; i >= 0; i--) {
      pos[i] = pos[i + 1] | v[i + 1];
    }
    long long maxx = -1;
    for (typeof(n) i = 0; i < n; i++) {
      if (i == 0) {
        maxx = max(maxx, pos[0] | (t * v[i]));
        continue;
      } else if (i == n - 1) {
        maxx = max(maxx, pre[i] | (t * v[i]));
        break;
      }
      maxx = max(maxx, pre[i] | pos[i] | (t * v[i]));
    }
    cout << maxx << endl;
  }
  return 0;
}

```