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 ll = long long;
int main() {
  ll n, k, x;
  cin >> n >> k >> x;
  vector<ll> as(n);
  for (auto&& a : as) cin >> a;
  ll kx = 1;
  for (int i = 0; i < k; ++i) kx *= x;
  vector<ll> or_front(n);
  or_front[0] = as[0];
  for (int i = 1; i < n; ++i) or_front[i] = as[i] | or_front[i - 1];
  vector<ll> or_back(n);
  or_back[n - 1] = as[n - 1];
  for (int i = n - 2; i >= 0; --i) or_back[i] = as[i] | or_back[i + 1];
  ll ans = 0;
  for (int i = 0; i < n; ++i) {
    const int front = i - 1 >= 0 ? or_front[i - 1] : 0;
    const int back = i + 1 < n ? or_back[i + 1] : 0;
    ans = max(ans, front | back | (kx * as[i]));
  }
  cout << ans << endl;
  return 0;
}
