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>
int main() {
  static int aa[200000], pp[200000], qq[200000];
  int n, k, i, p, q;
  long long x, xk, ans, a;
  scanf("%d%d%lld", &n, &k, &x);
  for (i = 0; i < n; i++) scanf("%d", &aa[i]);
  for (i = 0, p = 0; i < n; i++) {
    pp[i] = p;
    p |= aa[i];
  }
  for (i = n - 1, q = 0; i >= 0; i--) {
    qq[i] = q;
    q |= aa[i];
  }
  xk = 1;
  while (k--) xk *= x;
  ans = 0;
  for (i = 0; i < n; i++)
    if (ans < (a = pp[i] | aa[i] * xk | qq[i])) ans = a;
  printf("%lld\n", ans);
  return 0;
}
