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;
inline int getint() {
  static char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  int res = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') res = res * 10 + c - '0';
  return res;
}
inline void relax(long long &a, const long long &b) {
  if (b > a) a = b;
}
const int MaxN = 200005;
int n, k, x;
long long res = 0, a[MaxN];
long long pref[MaxN];
long long suff[MaxN];
int main() {
  n = getint(), k = getint(), x = getint();
  for (int i = 1; i <= n; ++i) a[i] = getint();
  long long m = 1;
  while (k--) m *= x;
  for (int i = 1; i <= n; ++i) pref[i] = pref[i - 1] | a[i];
  for (int i = n; i >= 1; --i) suff[i] = suff[i + 1] | a[i];
  for (int i = 1; i <= n; ++i) {
    long long w = pref[i - 1] | a[i] * m | suff[i + 1];
    relax(res, w);
  }
  cout << res;
  return 0;
}
