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;
inline unsigned long long power(unsigned long long base, int exp) {
  unsigned long long ans = 1;
  while (exp) {
    if (exp % 2 == 1) ans = (ans * base);
    base = (base * base);
    exp /= 2;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  unsigned long long n, k, x;
  cin >> n >> k >> x;
  unsigned long long *A = new unsigned long long[n + 1];
  for (int i = 0; i < n; ++i) cin >> A[i];
  sort(A, A + n);
  x = power(x, k);
  unsigned long long *pre = new unsigned long long[n + 1];
  pre[0] = A[0];
  for (int i = 1; i < n; ++i) pre[i] = pre[i - 1] | A[i];
  unsigned long long *suf = new unsigned long long[n + 1];
  suf[n] = 0;
  suf[n - 1] = A[n - 1];
  for (int i = n - 2; i >= 0; --i) suf[i] = suf[i + 1] | A[i];
  unsigned long long max0 = 0, temp;
  max0 = (A[0] * x) | suf[1];
  for (int i = 1; i < n - 1; ++i) {
    temp = pre[i - 1] | ((A[i] * x) | suf[i + 1]);
    max0 = max(max0, temp);
  }
  if (n != 1) max0 = max(max0, pre[n - 2] | (A[n - 1] * x));
  cout << max0 << endl;
  return 0;
}
