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;
long long a[1000000];
long long pre[1000000], post[1000000];
int main() {
  long long mx = 0;
  int n, k, x;
  scanf("%d", &n);
  scanf("%d", &k);
  scanf("%d", &x);
  for (int i = 0; i < n; i++) cin >> a[i];
  pre[0] = a[0];
  for (int i = 1; i < n; i++) pre[i] = pre[i - 1] | a[i];
  post[n - 1] = a[n - 1];
  for (int i = n - 2; i >= 0; i--) post[i] = post[i + 1] | a[i];
  for (int i = 0; i < n; i++) {
    long long tot = a[i];
    for (int j = 0; j < k; j++) tot *= x;
    if (i > 0) tot |= pre[i - 1];
    if (i < (n - 1)) tot |= post[i + 1];
    mx = max(tot, mx);
  }
  cout << mx << endl;
  return 0;
}
