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;
int gcd(int a, int b) {
  while (a && b) a > b ? a %= b : b %= a;
  return a + b;
}
int LCM(int a, int b) { return (a * b) / gcd(a, b); }
const int INF = INT_MAX;
const long long INFL = LLONG_MAX;
long long A[200009], prefix[200009], suffix[200009];
int main() {
  std::ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, k, x;
  cin >> n >> k >> x;
  long long num = 1;
  for (int(i) = 0; (i) < (k); (i)++) num *= x;
  for (int(i) = 0; (i) < (n); (i)++) cin >> A[i + 1];
  int i;
  for (i = 1; i <= n; i++) {
    prefix[i] = prefix[i - 1] | A[i];
  }
  for (i = n; i > 0; i--) suffix[i] = suffix[i + 1] | A[i];
  long long ans = 0;
  for (i = 1; i <= n; i++) {
    ans = max(ans, prefix[i - 1] | suffix[i + 1] | A[i] * num);
  }
  cout << ans << '\n';
  return 0;
}
