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 int64 = long long;
constexpr int DEBUG = 0;
template <class T>
inline bool UpdateMin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool UpdateMax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k, c;
  cin >> n >> k >> c;
  vector<int64> xs(n);
  for (int i = 0; i < n; i++) {
    cin >> xs[i];
  }
  vector<int64> ls(n), rs(n);
  ls[0] = xs[0];
  for (int i = 1; i < n; i++) {
    ls[i] = ls[i - 1] | xs[i];
  }
  rs[n - 1] = xs[n - 1];
  for (int i = n - 2; i >= 0; i--) {
    rs[i] = rs[i + 1] | xs[i];
  }
  int64 ans = 0;
  for (int i = 0; i < n; i++) {
    int64 x = xs[i];
    for (int j = 0; j < k; j++) {
      x *= c;
    }
    if (i - 1 >= 0) x |= ls[i - 1];
    if (i + 1 < n) x |= rs[i + 1];
    UpdateMax(ans, x);
  }
  cout << ans << endl;
}
