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;
int main() {
  int n, k, x;
  long long int tmul = 1, ans = 0;
  scanf("%d %d %d", &n, &k, &x);
  long long int a[n], left[n], right[n];
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  for (int i = 0; i < k; i++) tmul *= x;
  left[0] = a[0], right[n - 1] = a[n - 1];
  for (int i = 1; i < n; i++) left[i] = left[i - 1] | a[i];
  for (int i = n - 2; i >= 0; i--) right[i] = right[i + 1] | a[i];
  for (int i = 1; i < n - 1; i++) {
    ans = max(ans, (left[i - 1]) | right[i + 1] | (a[i] * tmul));
  }
  if (n == 1)
    ans = max(ans, (0 | 0 | (a[0] * tmul)));
  else {
    ans = max(ans, (0 | right[1] | (a[0] * tmul)));
    ans = max(ans, (left[n - 2]) | 0 | (a[n - 1] * tmul));
  }
  printf("%lld", ans);
  return 0;
}
