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;
unsigned long long n, k, x;
unsigned long long a[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  while (t--) {
    cin >> n >> k >> x;
    for (int i = 0; i < n; i++) cin >> a[i];
    unsigned long long pre[n + 1], pos[n + 1];
    unsigned long long res = 0;
    pre[0] = a[0];
    for (int i = 1; i < n; i++) {
      pre[i] = a[i];
      pre[i] |= pre[i - 1];
    }
    pos[n - 1] = a[n - 1];
    for (int i = n - 2; i >= 0; i--) {
      pos[i] = a[i];
      pos[i] |= pos[i + 1];
    }
    res = pre[n - 1];
    for (int i = 0; i < n; i++) {
      unsigned long long s = (i - 1) >= 0 ? pre[i - 1] : 0;
      unsigned long long r = (i + 1) < n ? pos[i + 1] : 0;
      unsigned long long tmp = a[i];
      for (int j = 0; j < k; j++) {
        tmp *= x;
        res = max(res, s | tmp | r);
      }
    }
    cout << res << endl;
  }
  return 0;
}
