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 oo = (int)1e9;
const double PI = 2 * acos(0.0);
const double eps = 1e-9;
const int MX = 200009;
int n, k, x;
int nums[MX];
int bin[MX][35];
int tot[35];
vector<int> tob(int x) {}
int main() {
  scanf("%d%d%d", &n, &k, &x);
  for (int i = 0; i < (int)(n); i++) {
    scanf("%d", &nums[i]);
    int t = nums[i];
    int ind = 0;
    while (t) {
      if (t & 1) {
        bin[i][ind] = 1;
        tot[ind]++;
      }
      t >>= 1;
      ind++;
    }
  }
  long long kx = 1;
  while (k--) kx *= x;
  long long ret = 0;
  for (int i = 0; i < (int)(n); i++) {
    long long t = 0;
    for (int j = 0; j < (int)(35); j++)
      if (tot[j] - bin[i][j]) t |= (1ll << j);
    t |= (kx * nums[i]);
    ret = max(ret, t);
  }
  cout << ret << endl;
  return 0;
}
