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 n, k, x, a[200020], b[200020], ans, i, j, v;
long long s[200020], t[200020];
int main() {
  scanf("%I64d%I64d%I64d", &n, &k, &x);
  for (i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  for (i = 1; i <= n; i++) {
    b[i] = a[i];
    for (j = 1; j <= k; j++) b[i] *= x;
  }
  for (i = 1; i <= n; i++)
    for (j = 0; j <= 62; j++)
      if ((1LL << j) & a[i]) s[j]++;
  for (i = 1; i <= n; i++) {
    v = 0;
    for (j = 0; j <= 62; j++) {
      if ((1LL << j) & a[i])
        t[j] = s[j] - 1;
      else
        t[j] = s[j];
      if (t[j]) v |= (1LL << j);
    }
    ans = max(ans, v | b[i]);
  }
  printf("%I64d\n", ans);
  return 0;
}
