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;
bool sortbysec(const pair<long long int, long long int>& a,
               const pair<long long int, long long int>& b) {
  return (a.second < b.second);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, k, x, i, p = 1;
  cin >> n >> k >> x;
  long long int arr[n], ma = 0;
  long long int pre[n + 1], suf[n + 1];
  pre[0] = 0;
  suf[n] = 0;
  for (i = 0; i < n; i++) cin >> arr[i];
  for (i = 1; i <= n; i++) pre[i] = pre[i - 1] | arr[i - 1];
  for (i = n - 1; i >= 0; i--) suf[i] = suf[i + 1] | arr[i];
  while (k--) p *= x;
  for (i = 0; i < n; i++) {
    long long int ans = 0;
    ans = (pre[i] | suf[i + 1] | p * arr[i]);
    ma = max(ma, ans);
  }
  cout << ma;
  return 0;
}
