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;
using namespace std;
long long a[30000000];
int main() {
  int n, k, x;
  long long pot = 1, todos = 0, unico = 0, res = 0;
  scanf("%d", &n);
  scanf("%d%d", &k, &x);
  for (int i = 0; i < k; i++) pot *= x;
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &a[i]);
    unico ^= (a[i] ^ (a[i] & (todos ^ unico)));
    todos |= a[i];
  }
  for (int i = 0; i < n; i++) {
    long long res2 = (a[i] & unico) ^ todos;
    res2 |= (a[i] * pot);
    res = max(res, res2);
  }
  printf("%I64d\n", res);
}
