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;
void smain();
int main() {
  ios_base::sync_with_stdio(0);
  smain();
  return 0;
}
long long n, k, x;
long long a[200001], l[200001], r[200001];
long long solve() {
  long long y = 1;
  for (long long i = 0; i < k; i++) y *= x;
  for (long long i = 0; i < n; i++) l[i] = (i ? l[i - 1] : 0) | a[i];
  for (long long i = n - 1; i >= 0; i--)
    r[i] = (i < n - 1 ? r[i + 1] : 0) | a[i];
  long long res = 0;
  for (long long i = 0; i < n; i++) {
    long long v = a[i] * y;
    v |= i ? l[i - 1] : 0;
    v |= i < n - 1 ? r[i + 1] : 0;
    res = max(res, v);
  }
  return res;
}
void smain() {
  for (; cin >> n >> k >> x;) {
    for (long long i = 0; i < n; i++) cin >> a[i];
    cout << solve() << '\n';
  }
}
