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;
void StartUp() {}
void StartUp(string in, string out) {
  freopen(in.c_str(), "r", stdin);
  freopen(out.c_str(), "w", stdout);
}
void StartUp(string str, string in, string out) {
  StartUp(str + "." + in, str + "." + out);
}
void StartUp(string str, bool def) {
  if (def) {
    StartUp(str, "in", "out");
  } else {
    StartUp(str, "gir", "cik");
  }
}
void StartUp(string str) { StartUp(str, false); }
template <typename T, typename U>
ostream& operator<<(ostream& out, pair<T, U>& v) {
  return out << "(" << v.first << "," << v.second << ")";
}
template <typename T>
void DBG(T x) {
  cerr << "x"
       << " =   " << x << endl;
}
template <typename T>
void DBG(T x, int n) {
  cerr << endl
       << "x"
       << " =" << endl;
  for (typeof(0) i = 0; i < n; i += 1) cerr << x[i] << endl;
}
template <typename T>
void DBG(T x, int n, bool tr) {
  if (tr)
    DBG(x, n);
  else {
    cerr << "x"
         << " =   ";
    for (typeof(0) i = 0; i < n; i += 1) cerr << x[i] << " ";
    cerr << endl;
  }
}
template <typename T>
void DBG(T x, int n, int m) {
  cerr << endl << "x" << endl;
  for (typeof(0) i = 0; i < n; i += 1) {
    for (typeof(0) j = 0; j < m; j += 1) {
      cerr << x[i][j] << " ";
    }
    cerr << endl;
  }
  cerr << endl;
}
long long v[(1 << 18)], n, m = 1, ans, dnums, nums;
long long calc(long long x) {
  long long t = ~(x & dnums);
  return t & nums;
}
int main() {
  StartUp();
  ios_base::sync_with_stdio(false);
  long long tmp, x;
  cin >> n >> tmp >> x;
  while (tmp--) m *= x;
  for (typeof(0) i = 0; i < n; i += 1) cin >> v[i];
  for (typeof(0) i = 0; i < n; i += 1) {
    dnums = dnums | (v[i] & nums);
    nums = nums | v[i];
  }
  dnums = ~dnums;
  for (typeof(0) i = 0; i < n; i += 1) {
    long long x = calc(v[i]);
    ans = max(ans, x | (v[i] * m));
  }
  cout << ans << endl;
}
