#include <bits/stdc++.h>
using namespace std;
inline int MAX(int a, int b) { return (a < b) ? b : a; }
inline int MIN(int a, int b) { return (a > b) ? b : a; }
inline int ABS(int a) { return (a > 0) ? a : -a; }
int n, k, x;
long long int a[200100];
pair<long long int, long long int> v[200100];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k >> x;
  for (int i = 0; i < n; ++i) cin >> a[i];
  long long int p = 1;
  for (int i = 0; i < k; ++i) p *= x;
  long long int sum = 0, value = 0;
  for (long long int i = 0, cur = 0; i < n; ++i) {
    v[i].first = cur;
    cur |= a[i];
  }
  for (long long int i = n - 1, cur = 0; i >= 0; --i) {
    v[i].second = cur;
    cur |= a[i];
  }
  for (int i = 0; i < n; ++i) {
    long long int sumaux = v[i].first | (a[i] * p) | v[i].second;
    if (sumaux > sum) {
      sum = sumaux;
    }
  }
  cout << sum << endl;
  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;
inline int MAX(int a, int b) { return (a < b) ? b : a; }
inline int MIN(int a, int b) { return (a > b) ? b : a; }
inline int ABS(int a) { return (a > 0) ? a : -a; }
int n, k, x;
long long int a[200100];
pair<long long int, long long int> v[200100];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k >> x;
  for (int i = 0; i < n; ++i) cin >> a[i];
  long long int p = 1;
  for (int i = 0; i < k; ++i) p *= x;
  long long int sum = 0, value = 0;
  for (long long int i = 0, cur = 0; i < n; ++i) {
    v[i].first = cur;
    cur |= a[i];
  }
  for (long long int i = n - 1, cur = 0; i >= 0; --i) {
    v[i].second = cur;
    cur |= a[i];
  }
  for (int i = 0; i < n; ++i) {
    long long int sumaux = v[i].first | (a[i] * p) | v[i].second;
    if (sumaux > sum) {
      sum = sumaux;
    }
  }
  cout << sum << endl;
  return 0;
}

```