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 main() {
  int n, k, x;
  cin >> n >> k >> x;
  vector<long long> ve;
  for (int i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    ve.push_back(x);
  }
  if (n == 1) {
    for (int i = 0; i < k; ++i) {
      ve[0] *= x;
    }
    cout << ve[0];
    return 0;
  }
  long long bestans = -1;
  vector<long long> pref(n), suf(n);
  for (int i = 0; i < n; ++i) {
    if (i - 1 >= 0)
      pref[i] = pref[i - 1] | ve[i];
    else
      pref[i] = ve[i];
    if (n - i < n)
      suf[n - i - 1] = suf[n - i] | ve[n - i - 1];
    else
      suf[n - i - 1] = ve[n - i - 1];
  }
  for (int i = 0; i < ve.size(); ++i) {
    long long newval = ve[i];
    for (int j = 0; j < k; ++j) {
      newval *= x;
    }
    long long ans;
    if (i == 0) {
      ans = suf[i + 1] | newval;
    } else if (i == ve.size() - 1) {
      ans = pref[ve.size() - 2] | newval;
    } else {
      ans = pref[i - 1] | suf[i + 1] | newval;
    }
    bestans = max(ans, bestans);
  }
  cout << bestans;
  return 0;
}
