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 long long mod = 998244353;
long long a[200006];
long long sparse[200006][20];
long long orin(long long lo, long long hi) {
  if (lo > hi) return 0;
  long long len = hi - lo + 1;
  while (len & len - 1) {
    len = len & len - 1;
  }
  len = log2(len);
  return sparse[lo][len] | sparse[hi - (1ll << len) + 1][len];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, x, k;
  cin >> n >> k >> x;
  for (long long i = (long long)0; i != (long long)n; i = i + 1) {
    cin >> a[i];
  }
  long long y = log2(n);
  for (long long i = (long long)0; i != (long long)n; i = i + 1) {
    sparse[i][0] = a[i];
  }
  for (long long j = (long long)1; j != (long long)y + 1; j = j + 1) {
    for (long long i = (long long)0; i != (long long)n; i = i + 1) {
      if (i + (1ll << j) > n) break;
      long long z = i + (1ll << j - 1);
      sparse[i][j] = sparse[i][j - 1] | sparse[z][j - 1];
    }
  }
  long long mu = pow(x, k), ans = 0;
  for (long long i = (long long)0; i != (long long)n; i = i + 1) {
    long long an = orin(0, i - 1);
    an = an | orin(i + 1, n - 1) | a[i] * mu;
    ans = max(ans, an);
  }
  cout << ans << '\n';
}
