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 or_left[200000];
long long or_right[200000];
long long pow_int(long long x, int m) {
  long long result = 1;
  for (int i = 1; i <= m; i++) result *= x;
  return result;
}
int main() {
  int n, k, x;
  scanf("%d %d %d", &n, &k, &x);
  vector<int> a;
  long long l = 0;
  for (int i = 0; i < n; i++) {
    int t;
    scanf("%d", &t);
    a.push_back(t);
    l = (l | t);
    or_left[i] = l;
  }
  long long r = 0;
  for (int i = n - 1; i >= 0; i--) {
    r = (r | a[i]);
    or_right[i] = r;
  }
  long long result = 0;
  for (int i = 0; i < n; i++) {
    long long orl = (i == 0 ? 0 : or_left[i - 1]);
    long long orr = (i == n - 1 ? 0 : or_right[i + 1]);
    long long result_here = ((orl | orr) | a[i] * pow_int(x, k));
    result = max(result_here, result);
  }
  cout << result;
  return 0;
}
