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;
int main() {
  int n, k, x;
  cin >> n >> k >> x;
  vector<long long> v(n), lft(n), rgt(n);
  long long cumm = 0;
  long long z = 1;
  while (k--) z *= x;
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
    cumm = lft[i] = cumm | v[i];
  }
  cumm = 0;
  for (int i = n - 1; i >= 0; i--) {
    cumm = rgt[i] = cumm | v[i];
  }
  long long ans = 0;
  auto f = [&](int i) {
    long long ret = 0;
    if (i + 1 < n) ret |= rgt[i + 1];
    if (i > 0) ret |= lft[i - 1];
    return ret;
  };
  for (int i = 0; i < n; i++) {
    long long cur = f(i);
    cur |= (z * v[i]);
    ans = max(ans, cur);
  }
  printf("%lld\n", ans);
  return 0;
}
