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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(9);
  int T = 1;
  while (T-- > 0) {
    int n, k, x;
    cin >> n >> k >> x;
    vector<int> a;
    vector<int> dig(32);
    int total = 0;
    for (int i = 0; i < n; ++i) {
      int x;
      cin >> x;
      total |= x;
      a.push_back(x);
      for (int j = 0; j < 32; ++j) {
        int mask = 1 << j;
        if (mask & x) {
          dig[j]++;
        }
      }
    }
    vector<int> b;
    for (int i = 0; i < n; ++i) {
      int res = total;
      for (int j = 0; j < 32; ++j) {
        if (dig[j]) {
          int mask = 1 << j;
          if (a[i] & mask) {
            if (dig[j] == 1) {
              res -= mask;
            }
          }
        }
      }
      b.push_back(res);
    }
    long long p = 1;
    for (int j = 0; j < k; ++j) p *= x;
    long long maxres = 0;
    for (int i = 0; i < n; ++i) {
      long long res = b[i] | (a[i] * 1ll * p);
      maxres = max(maxres, res);
    }
    cout << maxres << "\n";
  }
  return 0;
}
