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;
using pii = pair<int, int>;
const int N = 2e5;
int n, k, x;
int32_t a[N], t[4 * N];
int64_t xk = 1, ans;
void init(int i, int l, int r) {
  if (l == r) {
    t[i] = a[l];
  } else {
    int m = (l + r) / 2;
    init(2 * i + 1, l, m);
    init(2 * i + 2, m + 1, r);
    t[i] = t[2 * i + 1] | t[2 * i + 2];
  }
}
int64_t update(int i, int ui, int l, int r) {
  if (ui < l || r < ui) return t[i];
  if (l == r) return xk * t[i];
  int m = (l + r) / 2;
  return update(2 * i + 1, ui, l, m) | update(2 * i + 2, ui, m + 1, r);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout << fixed;
  cin >> n >> k >> x;
  for (int i = 0; i < n; i++) cin >> a[i];
  while (k-- > 0) xk *= x;
  init(0, 0, n - 1);
  for (int i = 0; i < n; i++) ans = max(ans, update(0, i, 0, n - 1));
  cout << ans << endl;
  return 0;
}
