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 tu[200001], qz[200002], hz[200002];
int main() {
  long long daan = 0;
  int n, k, x;
  cin >> n >> k >> x;
  for (int a = 1; a <= n; a++) scanf("%lld", &tu[a]);
  for (int a = 1; a <= n; a++) qz[a] = tu[a] | qz[a - 1];
  for (int a = n; a >= 1; a--) hz[a] = tu[a] | hz[a + 1];
  long long qw = 1;
  for (int a = 1; a <= k; a++) qw *= x;
  for (int a = 1; a <= n; a++)
    daan = max(daan, qz[a - 1] | hz[a + 1] | tu[a] * qw);
  cout << daan;
}
