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 MOD = 1e9 + 7;
const long long MX = 200010;
const long long INF = 1e9;
void print_a(vector<int> v) {
  if (v.size()) cout << v[0];
  for (int i = 1; i < v.size(); ++i) cout << ' ' << v[i];
  cout << '\n';
}
vector<vector<int> > init_vvi(int n, int m, int val) {
  return vector<vector<int> >(n, vector<int>(m, val));
}
vector<vector<long long> > init_vvl(int n, int m, long long val) {
  return vector<vector<long long> >(n, vector<long long>(m, val));
}
int pf[MX], sf[MX];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k, x;
  cin >> n >> k >> x;
  int a[n + 1];
  for (int i = 1; i < n + 1; ++i) {
    cin >> a[i];
  }
  long long ans = 0, m = 1;
  for (int i = 0; i < k; ++i) m = m * x;
  for (int i = 1; i < n + 1; ++i) pf[i] = pf[i - 1] | a[i];
  for (int i = n; i >= 1; --i) sf[i] = sf[i + 1] | a[i];
  for (int i = 1; i < n + 1; ++i) {
    ans = max(ans, pf[i - 1] | (a[i] * m) | sf[i + 1]);
  }
  cout << ans << '\n';
  return 0;
}
