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 long long max(long long a, long long b) { return a > b ? a : b; }
int a[200000 + 5], n, k, x;
int l[200000 + 5], r[200000 + 5];
long long ans;
long long Pow(long long n, int k) {
  long long res = 1;
  while (k) {
    if (k & 1) res *= n;
    n *= n;
    k >>= 1;
  }
  return res;
}
int main() {
  int i;
  scanf("%d%d%d", &n, &k, &x);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    l[i] = l[i - 1] | a[i];
  }
  for (i = n; i >= 1; i--) r[i] = r[i + 1] | a[i];
  for (i = 1; i <= n; i++)
    ans = max(ans, l[i - 1] | a[i] * Pow(x, k) | r[i + 1]);
  printf("%I64d\n", ans);
}
