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 rd() {
  int x = 0, ch = getchar(), f = 1;
  while (!isdigit(ch) && (ch != '-') && (ch != EOF)) ch = getchar();
  if (ch == '-') {
    f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void rt(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10)
    rt(x / 10), putchar(x % 10 + '0');
  else
    putchar(x + '0');
}
const int maxn = 200005;
int n, K, x;
long long a[maxn], b[maxn], c[maxn];
int main() {
  n = rd(), K = rd(), x = rd();
  for (int i = 1; i <= int(n); i++) a[i] = rd();
  for (int i = 1; i <= int(n); i++) b[i] = a[i] | b[i - 1];
  for (int i = n; i >= int(1); i--) c[i] = a[i] | c[i + 1];
  long long ans = 0, bs = 1;
  for (int i = 1; i <= int(K); i++) bs *= x;
  for (int i = 1; i <= int(n); i++)
    ans = max(ans, (a[i] * bs) | b[i - 1] | c[i + 1]);
  rt(ans);
}
