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;
unsigned long long a[1000111], b[1000011], forw[1000111];
int main() {
  int i, n, k, loc = -1;
  unsigned long long mx = 0LLU, x;
  scanf("%d %d %llu", &n, &k, &x);
  for (i = 0; i < n; i++) scanf("%llu", &a[i]);
  forw[0] = a[0];
  for (i = 1; i < n; i++) forw[i] = forw[i - 1] | a[i];
  b[n - 1] = a[n - 1];
  for (i = n - 2; i >= 0; i--) b[i] = b[i + 1] | a[i];
  b[n] = 0;
  unsigned long long ans;
  unsigned long long t = a[0];
  for (i = 0; i < k; i++) t = t * x;
  ans = t | b[1];
  for (i = 0; i < n; i++) {
    unsigned long long t = a[i];
    for (int j = 0; j < k; j++) t = t * x;
    unsigned long long y = forw[i - 1] | t;
    y = y | b[i + 1];
    ans = max(ans, y);
  }
  printf("%llu\n", ans);
  return 0;
}
