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 N = 200005;
const int M = 1000005;
const int MOD = 1000000007;
int n, k;
long long x;
long long arr[N], pref[N], suff[N];
int main() {
  scanf("%d%d%lld", &n, &k, &x);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", arr + i);
    pref[i] = arr[i] | pref[i - 1];
  }
  for (int i = n; i > 0; --i) suff[i] = arr[i] | suff[i + 1];
  long long aux = 1;
  while (k--) aux *= x;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    ans = max(ans, pref[i - 1] | (arr[i] * aux) | suff[i + 1]);
  }
  printf("%lld\n", ans);
  return 0;
}
