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;
const int MN = 100 * 100 * 30 + 100;
long long k, n, x, a[MN], b[MN], pw[22], r[MN], l[MN], mul = 1;
int main() {
  ios ::sync_with_stdio(false);
  pw[0] = 1;
  for (int i = 1; i < 22; i++) pw[i] = pw[i - 1] * 2;
  cin >> n >> k >> x;
  while (k--) mul *= x;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
    l[i] = a[i];
    b[i] *= mul;
  }
  for (int i = n - 1; i >= 0; i--) r[i] = a[i] | r[i + 1];
  for (int i = 1; i < n; i++) l[i] = a[i] | l[i - 1];
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    ans = max(ans, (l[i - 1] | b[i] | r[i + 1]));
  }
  cout << ans;
  return 0;
}
