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.
#include <bits/stdc++.h>
using namespace std;
template <class T>
void print_array(T a[], int size) {
  for (int i = 0; i < size; i++) cout << a[i] << " ";
  cout << endl;
}
template <class T>
void print_array_v(T &a) {
  int size = a.size();
  for (int i = 0; i < size; i++) cout << a[i] << " ";
  cout << endl;
}
int main() {
  int a[200200];
  int b[200200] = {0};
  int cnt[50] = {0};
  int n, k, x;
  scanf("%d", &n);
  scanf("%d", &k);
  scanf("%d", &x);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i <= 30; i++)
    for (int j = 0; j < n; j++)
      if ((a[j] & (1 << i)) != 0) cnt[i]++;
  for (int i = 0; i <= 30; i++)
    for (int j = 0; j < n; j++)
      if (cnt[i] - ((a[j] & (1 << i)) != 0) > 0) b[j] |= (1 << i);
  long long ans = 0;
  long long m = 1;
  for (int i = 0; i < k; i++) m *= x;
  for (int i = 0; i < n; i++) ans = max(ans, b[i] | (a[i] * m));
  cout << ans << endl;
  return 0;
}
