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;
static long long check(long long x) {
  long long count = 0;
  while (x > 0) {
    count++;
    x /= 2;
  }
  return count;
}
int main() {
  long long n, k, x;
  cin >> n >> k >> x;
  vector<long long> val(n);
  for (long long i = 0; i < n; i++) cin >> val[i];
  sort(val.begin(), val.end());
  long long x2 = x;
  k--;
  while (k > 0) {
    x *= x2;
    k--;
  }
  long long len = check(val[n - 1] * x);
  vector<long long> z(len);
  k = 1;
  for (long long i = 0; i < len; i++) {
    for (long long j = 0; j < n; j++)
      if ((val[j] / k) % 2 == 1) z[i]++;
    k *= 2;
  }
  long long max = 0;
  for (long long i = n - 1; i > -1; i--) {
    long long temp = x * val[i];
    k = 1;
    long long sum = 0;
    for (long long j = 0; j < len; j++) {
      if (z[j] > 1 || (temp / k) % 2 == 1 ||
          (z[j] == 1 && (val[i] / k) % 2 == 0))
        sum += k;
      k *= 2;
    }
    if (sum > max) max = sum;
  }
  cout << max << endl;
  return 0;
}
