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 n, k, x;
long long res, r2;
vector<long long> V, V2;
long long P[2][200002];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k >> x;
  long long a = 1;
  for (int i = 0; i < k; i++) a *= x;
  for (int i = 0; i < n; i++) {
    int b;
    cin >> b;
    V.push_back(b);
    V2.push_back(b * a);
  }
  for (int i = 0; i < (int)V.size(); i++) {
    res |= V[i];
    P[0][i] = res;
  }
  res = 0;
  for (int i = (int)V.size() - 1; i >= 0; i--) {
    res |= V[i];
    P[1][i] = res;
  }
  for (int i = 0; i < (int)V.size(); i++) {
    long long tmp;
    if (i == 0)
      tmp = V2[i] | P[1][i + 1];
    else
      tmp = (P[0][i - 1] | V2[i]) | P[1][i + 1];
    if (tmp > r2) r2 = tmp;
  }
  cout << r2 << '\n';
  return 0;
}
