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;
long long a[200005], b[200005], pre[200005], suf[200005];
int m;
long long ci;
int main() {
  int n, k, x;
  scanf("%d%d%d", &n, &k, &x);
  ci = 1;
  for (int i = 1; i <= k; i++) ci = ci * x;
  long long mx = 0;
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &a[i]);
  }
  pre[0] = 0;
  for (int i = 1; i < n; i++) pre[i] = pre[i - 1] | a[i - 1];
  suf[n - 1] = 0;
  for (int i = n - 2; i >= 0; i--) suf[i] = suf[i + 1] | a[i + 1];
  for (int i = 0; i < n; i++) mx = max(mx, pre[i] | (a[i] * ci) | suf[i]);
  printf("%I64d\n", mx);
  return 0;
}
