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 inf = 1000000000ll;
const long long inf64 = inf * inf;
const long long base = inf + 7;
bool solve() {
  int n, k;
  unsigned long long x;
  cin >> n >> k >> x;
  vector<unsigned long long> a(n);
  for (int y, i = 0; i < n; i++) {
    scanf("%d", &y);
    a[i] = y;
  }
  sort((a).begin(), (a).end());
  vector<unsigned long long> preff(n), suff(n);
  unsigned long long t = 0;
  for (int i = 0; i < n; i++) {
    t |= a[i];
    preff[i] = t;
  }
  t = 0;
  for (int i = n - 1; i >= 0; i--) {
    t |= a[i];
    suff[i] = t;
  }
  unsigned long long ans = 0;
  for (int i = 0; i < n; i++) {
    t = 0;
    if (i > 0) t |= preff[i - 1];
    if (i + 1 < n) t |= suff[i + 1];
    unsigned long long v = a[i];
    for (int j = 0; j < k; j++) v *= x;
    t |= v;
    ans = max(ans, t);
  }
  cout << ans << '\n';
  return true;
}
int main() {
  solve();
  return 0;
}
