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 pow(long long g, long long e) {
  if (!e)
    return 1;
  else {
    long long ret = pow(g, e >> 1);
    ret *= ret;
    if ((e & 1)) ret *= g;
    return ret;
  }
}
int main() {
  iostream::sync_with_stdio(false);
  cin.tie(0);
  long long N, K, x;
  cin >> N >> K >> x;
  x = pow(x, K);
  long long a[200100];
  vector<vector<long long> > oc(40);
  for (long long n = 0; n < N; n++) {
    cin >> a[n];
    for (long long i = 31; i >= 0; i--) {
      if (((1 << i) & a[n])) {
        oc[i].push_back(n);
      }
    }
  }
  long long mx = 0;
  for (long long n = 0; n < N; n++) {
    long long cur = a[n] * x;
    for (long long i = 0; i < oc.size(); i++) {
      if (oc[i].size() > 1 || (oc[i].size() == 1 && oc[i][0] != n))
        cur |= (1ll << i);
    }
    mx = max(mx, cur);
  }
  cout << mx << endl;
  return 0;
}
