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 INF = 1e9;
const long long INF_LL = 4e18;
const double pi = acos(-1.0);
template <class T>
T& smin(T& a, const T& b) {
  if (a > b) a = b;
  return a;
}
template <class T>
T& smax(T& a, const T& b) {
  if (a < b) a = b;
  return a;
}
template <class T>
T sqr(const T& a) {
  return a * a;
}
template <typename Iterator>
void print(const char* fmt, Iterator begin, Iterator end,
           const char* separator = " ", const char* closing = "\n") {
  for (Iterator it = begin; it != end;
       printf(fmt, *it), printf("%s", ++it == end ? closing : separator))
    ;
}
const int N = 2e5 + 5;
int n, k, x, a[N];
long long l[N], r[N];
int main() {
  cin.tie(NULL);
  scanf("%d%d%d", &n, &k, &x);
  long long xp = 1;
  for (int i(0), _b(k); i < _b; ++i) xp *= x;
  for (int i(0), _b(n); i < _b; ++i) scanf("%d", a + i);
  l[0] = 0;
  for (int i(0), _b(n); i < _b; ++i) l[i + 1] = l[i] | a[i];
  r[n] = 0;
  for (int i(n - 1), _b(0); i >= _b; --i) r[i] = r[i + 1] | a[i];
  long long ans = l[n];
  for (int i(0), _b(n); i < _b; ++i) {
    long long t = l[i] | r[i + 1];
    smax(ans, t | (a[i] * xp));
  }
  cout << ans << endl;
  return 0;
}
