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 n, n1, k, x, i = 0, m, su = 0, su1, su2, I, maximo = 0;
vector<int> a, b, c;
int main() {
  cin >> n >> k >> x;
  i = 0;
  while (i < n) {
    cin >> m;
    a.push_back(m);
    i++;
  }
  i = 0;
  n1 = n - 1;
  b.push_back(a[0]);
  c.push_back(a[n1]);
  while (i < (n - 2)) {
    b.push_back(b[i] | a[i + 1]);
    c.push_back(c[i] | a[n1 - i - 1]);
    i++;
  }
  i = 0;
  n1 = n - 1;
  while (i < n) {
    su = 0;
    su1 = 0;
    su2 = a[i] * pow(x, k);
    if (i != 0) su = b[i - 1];
    if (i != n1) su1 = c[n1 - i - 1];
    maximo = max(maximo, (su | su1 | su2));
    i++;
  }
  cout << maximo;
  return 0;
}
