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 sigma_size = 26;
const int N = 100 + 50;
const int MAXN = 200000 + 50;
const int inf = 0x3fffffff;
const double eps = 1e-8;
const int HASH = 100007;
const int mod = 1000000000 + 7;
template <class T>
inline bool RD(T& res) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  res = (c == '-') ? 0 : c - '0';
  while (c = getchar(), c >= '0' && c <= '9') res = res * 10 + (c - '0');
  res *= sgn;
  return 1;
}
template <class T>
inline void PT(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) PT(x / 10);
  putchar(x % 10 + '0');
}
int n, k, x;
long long a[MAXN], l[MAXN], r[MAXN];
int main() {
  RD(n);
  RD(k);
  RD(x);
  long long num = 1;
  for (int i = (int)1; i <= (int)k; i++) num *= x;
  for (int i = (int)1; i <= (int)n; i++) RD(a[i]);
  l[0] = r[n + 1] = 0;
  for (int i = (int)1; i <= (int)n; i++) l[i] = l[i - 1] | a[i];
  for (int i = (int)n; i >= (int)1; i--) r[i] = r[i + 1] | a[i];
  long long ans = 0;
  for (int i = (int)1; i <= (int)n; i++)
    ans = max(ans, (l[i - 1] | a[i] * num) | r[i + 1]);
  printf("%lld\n", ans);
  return 0;
}
