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;
const int INF = 1 << 30;
unsigned long long n, k, x;
unsigned long long a[300000];
unsigned long long dpL[300000];
unsigned long long dpR[300000];
int main() {
  int i, j;
  cin >> n >> k >> x;
  unsigned long long s = 0, MAX = 0;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (i = 0; i < n - 1; i++) {
    dpL[i + 1] = dpL[i] | a[i];
  }
  for (i = n - 1; i > 0; i--) {
    dpR[i - 1] = dpR[i] | a[i];
  }
  unsigned long long ans1 = a[0], ans2 = a[n - 1];
  for (i = 0; i < k; i++) {
    ans1 = ans1 * x;
    ans2 = ans2 * x;
  }
  unsigned long long ans = max(ans1 | dpR[0], ans2 | dpL[n - 1]);
  for (i = 1; i < n - 1; i++) {
    unsigned long long ANS = a[i];
    for (j = 0; j < k; j++) {
      ANS *= x;
    }
    ans = max(ans, ANS | dpL[i] | dpR[i]);
  }
  cout << ans << endl;
  return 0;
}
