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 INF = 1000 * 1000 * 1000;
const long long INF64 = 1LL * INF * INF;
const long long mod = INF + 7;
long long a[200001];
long long pref[200001], post[200001];
void solve() {
  int n, k, x;
  cin >> n >> k >> x;
  int d = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; ++i) {
    pref[i] = pref[i - 1] | a[i - 1];
  }
  for (int i = n - 1; i >= 0; --i) {
    post[i] = post[i + 1] | a[i];
  }
  long long max = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= k; ++j) {
      long long tx = 1;
      for (int s = 0; s <= j; ++s) {
        max = std::max(max, (pref[i - 1] | (a[i - 1] * tx) | post[i]));
        tx *= x;
      }
    }
  }
  cout << max << endl;
}
int main() {
  cout.sync_with_stdio(false);
  solve();
  return 0;
}
