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;
long long factorial(long long n) {
  if (n == 0) return 1;
  return (n * factorial(n - 1)) % 1000000007;
}
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return (power(x, y / 2) * power(x, y / 2));
  else
    return (x * ((power(x, y / 2) * power(x, y / 2))));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tt = 1;
  while (tt--) {
    long long n, k, x;
    cin >> n >> k >> x;
    long long l[n];
    long long prefix[n];
    long long suffix[n];
    long long maxAns = 0;
    for (int i = 0; i < n; i++) {
      cin >> l[i];
    }
    if (n == 1) {
      cout << l[0] * power(x, k) << endl;
      return 0;
    }
    prefix[0] = l[0];
    for (int i = 1; i < n; i++) {
      prefix[i] = prefix[i - 1] | l[i];
    }
    suffix[n - 1] = l[n - 1];
    for (int i = n - 2; i >= 0; i--) {
      suffix[i] = suffix[i + 1] | l[i];
    }
    for (int i = 0; i < n; i++) {
      if (i == 0) {
        maxAns = max(maxAns, (l[i] * power(x, k)) | suffix[i + 1]);
        continue;
      }
      if (i == n - 1) {
        maxAns = max(maxAns, (l[i] * power(x, k)) | prefix[i - 1]);
        continue;
      } else {
        maxAns =
            max(maxAns, (l[i] * power(x, k)) | suffix[i + 1] | prefix[i - 1]);
      }
    }
    cout << maxAns << endl;
  }
}
