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 numbers[200002], leftt[200002], rightt[200002], x;
int n, k, imx;
void Read() {
  cin >> n >> k >> x;
  imx = -1;
  long long mx = -1;
  for (int i = 0; i < n; ++i) {
    cin >> numbers[i];
    if (i) {
      leftt[i] = leftt[i - 1] | numbers[i - 1];
    }
  }
  for (int i = n - 1; i >= 0; --i) {
    rightt[i] = rightt[i + 1] | numbers[i + 1];
  }
}
void Solve() {
  long long orsum = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < k; ++j) {
      numbers[i] *= x;
    }
  }
  for (int i = 0; i < n; ++i) {
    orsum = max(orsum, leftt[i] | numbers[i] | rightt[i]);
  }
  cout << orsum << "\n";
}
int main() {
  Read();
  Solve();
  return 0;
}
