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 long double pi = 3.1415926535897932384626433832795l;
template <typename T>
inline auto sqr(T x) -> decltype(x * x) {
  return x * x;
}
template <typename T1, typename T2>
inline bool umx(T1& a, T2 b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
inline bool umn(T1& a, T2 b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
const int K = 60;
const int N = 200100;
struct Input {
  int n, k, x;
  int a[N];
  bool read() {
    if (scanf("%d%d%d", &n, &k, &x) != 3) return 0;
    for (int i = int(0); i < int(n); ++i) {
      scanf("%d", &a[i]);
    }
    return 1;
  }
  void init(const Input& input) { *this = input; }
};
struct Data : Input {
  long long ans;
  void write() { cout << ans << endl; }
  virtual void solve() {}
  virtual void clear() { *this = Data(); }
};
struct Solution : Data {
  int cnt[K];
  int cnt1[K];
  void solve() {
    memset(cnt, 0, sizeof(cnt));
    for (int i = int(0); i < int(n); ++i) {
      for (int j = int(0); j < int(K); ++j) {
        if (a[i] & (1ll << j)) {
          cnt[j]++;
        }
      }
    }
    ans = 0;
    for (int i = int(0); i < int(n); ++i) {
      memcpy(cnt1, cnt, K * sizeof(int));
      for (int j = int(0); j < int(K); ++j) {
        if (a[i] & (1ll << j)) {
          cnt1[j]--;
        }
      }
      long long help = a[i];
      for (int j = int(0); j < int(k); ++j) {
        help *= x;
      }
      for (int j = int(0); j < int(K); ++j) {
        if (help & (1ll << j)) {
          cnt1[j]++;
        }
      }
      long long res = 0;
      for (int j = int(0); j < int(K); ++j) {
        if (cnt1[j]) {
          res |= (1ll << j);
        }
      }
      umx(ans, res);
    }
  }
  void clear() { *this = Solution(); }
};
Solution sol;
int main() {
  cout.setf(ios::showpoint | ios::fixed);
  cout.precision(20);
  while (sol.read()) {
    sol.solve();
    sol.write();
    sol.clear();
  }
  return 0;
}
