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 int maxn = 2e5 + 1;
int a[maxn];
long long prefix[maxn], suf[maxn];
class Solution {
 public:
  void solve(std::istream& in, std::ostream& out) {
    int n, k, x;
    in >> n >> k >> x;
    for (int i = 1; i <= n; i++) {
      in >> a[i];
    }
    long long m = 1;
    while (k--) {
      m *= x;
    }
    for (int i = 1; i <= n; i++) {
      prefix[i] = prefix[i - 1] | a[i];
    }
    for (int i = n; i >= 1; i--) {
      suf[i] = suf[i + 1] | a[i];
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      ans = max(ans, prefix[i - 1] | a[i] * m | suf[i + 1]);
    }
    out << ans;
  }
};
void solve(std::istream& in, std::ostream& out) {
  out << std::setprecision(12);
  Solution solution;
  solution.solve(in, out);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  istream& in = cin;
  ostream& out = cout;
  solve(in, out);
  return 0;
}
