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;
int main(void) {
  long long int i, k, x, m, n;
  cin >> n >> k >> x;
  long long int a[n], b[n], c[n];
  for (i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
    c[i] = a[i];
  }
  for (i = 1; i < n; i++) b[i] = b[i] | b[i - 1];
  for (i = (n - 2); i >= 0; i--) c[i] = c[i] | c[i + 1];
  m = b[n - 1];
  if (n == 1) m *= (long long int)pow(x, k);
  if (n > 1) {
    m = max(m, (b[0] * (long long int)pow(x, k)) | c[1]);
    m = max(m, (c[n - 1] * (long long int)pow(x, k)) | b[n - 2]);
  }
  for (i = 1; i < (n - 1); i++)
    m = max(m, b[i - 1] | (a[i] * (long long int)pow(x, k)) | c[i + 1]);
  cout << m;
  return 0;
}
