#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9, mod = 1e9 + 7, N = 1e5 + 9;
long long n, k, x, ans = 0, a[200000], f[200000];
long long pw(long long a, long long b) {
  int res = 1;
  for (int i = 1; i < b + 1; i++) res *= a;
  return res;
}
long long L[200000], R[200000];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k >> x;
  long long T = pw(x, k), ans = 0;
  for (int i = 0; i < n; i++) cin >> a[i];
  L[0] = a[0];
  for (int i = 1; i < n; i++) L[i] = L[i - 1] | a[i];
  R[n - 1] = a[n - 1];
  for (int i = n - 2; i >= 0; i--) R[i] = R[i + 1] | a[i];
  ans = max(L[n - 2] | (a[n - 1] * T), (R[1] | (a[0] * T)));
  for (int i = 1; i < n - 1; i++) {
    ans = max(ans, (L[i - 1] | (a[i] * T) | R[i + 1]));
  }
  cout << ans << "\n";
}

### Prompt

Please create a solution in cpp 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 INF = 1e9, mod = 1e9 + 7, N = 1e5 + 9;
long long n, k, x, ans = 0, a[200000], f[200000];
long long pw(long long a, long long b) {
  int res = 1;
  for (int i = 1; i < b + 1; i++) res *= a;
  return res;
}
long long L[200000], R[200000];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k >> x;
  long long T = pw(x, k), ans = 0;
  for (int i = 0; i < n; i++) cin >> a[i];
  L[0] = a[0];
  for (int i = 1; i < n; i++) L[i] = L[i - 1] | a[i];
  R[n - 1] = a[n - 1];
  for (int i = n - 2; i >= 0; i--) R[i] = R[i + 1] | a[i];
  ans = max(L[n - 2] | (a[n - 1] * T), (R[1] | (a[0] * T)));
  for (int i = 1; i < n - 1; i++) {
    ans = max(ans, (L[i - 1] | (a[i] * T) | R[i + 1]));
  }
  cout << ans << "\n";
}

```