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 N = 200500;
long long a[N];
long long p[N], s[N];
int n, k, x;
long long get(int i) {
  long long res = 0;
  if (i) res |= p[i - 1];
  if (i != n - 1) res |= s[i + 1];
  return res | (a[i] * x);
}
int main() {
  cin >> n >> k >> x;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i == 0)
      p[i] = a[i];
    else
      p[i] = p[i - 1] | a[i];
  }
  long long tmp = 1;
  for (int i = 0; i < k; i++) tmp *= x;
  x = tmp;
  long long res = 0;
  s[n - 1] = a[n - 1];
  for (int i = n - 2; i >= 0; i--) s[i] = s[i + 1] | a[i];
  int id = -1;
  for (int j = 0; j < n; j++)
    if (res <= get(j)) res = get(j), id = j;
  a[id] *= x;
  cout << res;
  return 0;
}
