#include <bits/stdc++.h>
const double Pi = 3.1415926;
using namespace std;
int main() {
  int n, k, x;
  cin >> n >> k >> x;
  vector<int> v(n);
  vector<set<int> > bit(30);
  long long int t, m, s;
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
    t = v[i];
    m = 0;
    while (t) {
      if (t & 1) {
        bit[m].insert(i);
      }
      t >>= 1;
      m++;
    }
  }
  m = 0;
  for (int i = 0; i < n; ++i) {
    s = v[i];
    for (int j = 0; j < k; ++j) s *= x;
    t = 1;
    for (int j = 0; j < 30; ++j) {
      if (bit[j].size() > 1 || (bit[j].size() == 1 && bit[j].count(i) == 0))
        s |= t;
      t <<= 1;
    }
    m = max(m, s);
  }
  cout << m;
  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>
const double Pi = 3.1415926;
using namespace std;
int main() {
  int n, k, x;
  cin >> n >> k >> x;
  vector<int> v(n);
  vector<set<int> > bit(30);
  long long int t, m, s;
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
    t = v[i];
    m = 0;
    while (t) {
      if (t & 1) {
        bit[m].insert(i);
      }
      t >>= 1;
      m++;
    }
  }
  m = 0;
  for (int i = 0; i < n; ++i) {
    s = v[i];
    for (int j = 0; j < k; ++j) s *= x;
    t = 1;
    for (int j = 0; j < 30; ++j) {
      if (bit[j].size() > 1 || (bit[j].size() == 1 && bit[j].count(i) == 0))
        s |= t;
      t <<= 1;
    }
    m = max(m, s);
  }
  cout << m;
  return 0;
}

```