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;
template <class T>
T min3(T a, T b, T c) {
  return min(a, min(b, c));
};
template <class T>
T max3(T a, T b, T c) {
  return max(a, max(b, c));
};
void openFile() { freopen("/home/khaihanhdk/devplace/a.inp", "r", stdin); }
int n, k;
long long x;
long long a[200100], b[200100], c[200100];
int main() {
  scanf("%d%d%I64d", &n, &k, &x);
  long long y = 1;
  for (int i = 0; i < k; i++) y *= x;
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  b[0] = 0;
  for (int i = 1; i <= n; i++) b[i] = a[i] | b[i - 1];
  c[n + 1] = 0;
  for (int i = n; i >= 1; i--) c[i] = a[i] | c[i + 1];
  long long mmax = 0;
  for (int i = 1; i <= n; i++) {
    mmax = max(mmax, (a[i] * y) | b[i - 1] | c[i + 1]);
  }
  mmax = max(mmax, a[1] | c[2]);
  cout << mmax;
}
